Please show all the work

1- How many grams are in 12.3 moles of Dinitrogen Pentoxide?

2-How many grams are in 2.7 moles of Iron (III) Nitrate? (Fe(NO3)3)

3- How many grams are in 0.16 moles of Sucrose? (C12H22011)

4- How many grams are in 0.87 moles of Potassium Iodide? (KI)

Please Show All The Work 1- How Many Grams Are In 12.3 Moles Of Dinitrogen Pentoxide? 2-How Many Grams

Answers

Answer 1

We haver the mass from the moles of each of the compounds below.

Number of moles

1) Mass/Molar mass = number of moles

=Mass = Molar Mass * Molar Number

= 12.3 moles x g/mol 108

= 1328.4 g

2) Mass/Molar mass = number of moles

=Mass = Molar Mass * Molar Number

2.7 moles times 242 g/mol

= 653 g

3)) Molar mass divided by the number of moles

=Mass = Molar Mass * Molar Number

= 3.42 g/mol * 0.16 moles

= 54.72 g

4) Moles = mass / molecular mass

=Mass = Molar Mass * Molar Number

= 166 g/mol * 0.87 moles

= 144 g

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Related Questions

A 346.9 mL sample of carbon dioxide was heated to 373 K. If the volume of the carbon dioxide sample at 373 K is 596.2 mL,
what was its temperature at 346.9 mL?
T =

Answers

Answer:

216.9 K

Explanation:

We can use the combined gas law to solve this problem:

(P1V1/T1) = (P2V2/T2)

where P is pressure, V is volume, and T is temperature. We can assume that the pressure is constant, since the problem doesn't mention any changes in pressure.

Let's label the initial temperature T1 and the final temperature T2, and plug in the given values:

(P1V1/T1) = (P2V2/T2)

V1 = 346.9 mL

V2 = 596.2 mL

T2 = 373 K

(P1 * 346.9 mL / T1) = (P1 * 596.2 mL / 373 K)

Simplifying and solving for T1:

T1 = (P1 * 346.9 mL * 373 K) / (P1 * 596.2 mL)

T1 = (346.9 * 373) / 596.2

T1 = 216.9 K

Therefore, the temperature of the carbon dioxide sample at 346.9 mL was 216.9 K.

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Select the correct answer. A certain reaction has this form:aA bB. At a particular temperature and [A]0 = 2.00 x 10-2 Molar, concentration versus time data were collected for this reaction and a plot of ln[A]t versus time resulted in a straight line with a slope value of -2.97 x 10-2 min-1. What is the reaction order and rate law for this reaction? A. first, rate = k[A] B. first, rate = k[A]2 C. second, rate = k[A] D. second, rate = k[A]2 E. third, rate = k[A]

Answers

The reaction order and rate law for this reaction is first, rate = k[A]. Option A is correct.

The reaction order and rate law for a reaction can be determined from the slope of a plot of ln[A] versus time.

Given that the plot of ln[A] versus time resulted in a straight line with a slope value of -2.97 x 10⁻² min⁻¹, we can determine the reaction order and rate law as follows;

If the slope is equal to 1, the reaction order is 1st order.

If the slope is equal to 2, the reaction order is 2nd order.

If the slope is equal to 3, the reaction order is 3rd order.

From the given slope of -2.97 x 10⁻² min⁻¹, we can conclude that the reaction order is 1st order, because the slope value is equal to -1 times the reaction order. So, the reaction order for this reaction is 1st order.

The rate law for a 1st order reaction will be given by;

rate = k[A]

where [A] is the concentration of the reactant, and k is the rate constant.

Therefore,  first, rate = k[A].

Hence, A. is the correct option.

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Give the number of significant figures indicated.

0.066

Answers

There are 3 significant figures, because after a . A 0 and the numbers after it count as sig figs

Watch the animation and select the interactions that can be explained by hydrogen bonding.
HF
is a weak acid neutralized by NaOH
.
HF
has a higher boiling point than HCl
.
CH4
molecules interact more closely in the liquid than in the gas phase.
Ice, H2O
, has a solid structure with alternating H−O
interactions.
H2Te
has a higher boiling point than H2S

Answers

The interactions that can be explained by hydrogen bonding are:

HF has a higher boiling point than HClIce, H₂O, has a solid structure with alternating H−O interactions.

Hydrogen bonding is a type of intermolecular force that occurs between a hydrogen atom in a polar molecule and a highly electronegative atom, such as nitrogen, oxygen, or fluorine, in a neighboring molecule. It is a relatively strong force and can significantly affect the physical and chemical properties of substances.

In the case of HF and HCl, both molecules are polar, but HF has a higher boiling point due to the stronger hydrogen bonding between its molecules. In ice, the hydrogen bonding between water molecules creates a crystal structure with a characteristic lattice arrangement.

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Calculate the mass of water produced when 1.03 g
of butane reacts with excess oxygen.

Answers

1.03 g of butane will yield 1.59 g of water when it reacts with too much oxygen.

What is mass?

The amount of matter in an item is measured by the fundamental physical quantity known as mass. It is a scalar quantity that is measured in grams (g) or kilograms (kg) (g). No matter where it is or what force is pushing against it, an object's mass always remains constant.

How do you determine it?

For butane and oxygen to burn, the chemical equation is balanced as follows:

2C4H10+13 O2→ 8 CO2+ 10 H2O

According to the equation, 1 mole of butane (C4H10) and 13/2 moles of oxygen (O2) combine to form 5 moles of water (H2O).

To begin with, we must count the butane moles that are present:

Mass of butane divided by its molar mass yields moles of butane.

1.03 g/58.12 g/mol is the formula for butane.

A mole of butane weighs 0.0177 mol.

Then, we can calculate the quantity of water created using the butane-to-water mole ratio:

Moles of water = [tex]\frac{5 mol H2O}{(1 mol C4H10 ×0.0177 mol C4H10)}[/tex] = 0.0885 mol.

Eventually, we can determine how much water was generated:

moles of water equal 0.0885 mol when 5 mol H2O is divided by 1 mol C4H10 and multiplied by 0.0177 mol C4H10.

Lastly, we can determine the mass of created water:

Water's mass is equal to its moles times its molar mass= 0.0885 mol × 18.02 g/mol = 1.59 g.

As a result, 1.03 g of butane will yield 1.59 g of water when it reacts with to moles of water equal 0.0885 mol when 5 mol H2O is divided by 1 mol C4H10 and multiplied by 0.0177 mol C4H10.

Lastly, we can determine the mass of created water:

Water's mass is equal to its moles times its molar mass = 0.0885 mol × 18.02 g/mol = 1.59 g.

As a result, 1.03 g of butane will yield 1.59 g of water when it reacts with too much oxygen.

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Explain to me please????​

Answers

Answer:Non polar.

Explanation:because water is polar because of its shape

Show that the following reaction obeys zero order kinetics, and find the rate constant and half-life. 10²x[A]/mol dm-3 13.80 11.05 8.43 5.69 3.00 Time/hr 0 10 20 30 40

Answers

The half life for the given reaction is 20s. The rate constant for the given zero order reaction can be given as 0.5mol L⁻¹s⁻¹.

According to the definition of a zero-order reaction, it is "a chemical reaction that occurs when the rate of response remains unchanged whether the percentage of the reactants increases or decreases." Since the rate increases according in proportion to the 0th magnitude of the reactant concentration, the rate at which these responses is always equivalent to the constant rate for the particular reactions.

k = 0.5mol L⁻¹s⁻¹

Half life is a/2K=3/2×40

                        = 20s

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What quantity of heat (in J) would be required to convert 0.27 mol of a pure substance from a liquid at 50 °C to a gas at 113.0 °C?.
Cliquid = 1.45 J/mol C
Cgas = 0.65 J/mol *C
Tboiling = 88.5 °C
AHvaporization = 1.23 kJ/mol
Give your answer in Joules

Answers

Explanation:

First you have to heat the liquid from 50 to 88.5 C  to get it boiling...

  then you need to boil it all to a gas....the you have to heat it to 113 C

.27 mole  *

     (  (88.5 -50 C)*1.45 J / (mole-C)      <====heating the liquid

               +  1230 J / mole                              <====boiling to a gas

                       + (113 - 88.5 C) * .65 J/(mole C)    <=====heating the gas

                                = 351 .5 J                              <=====result

Compare and contrast the four different types of solar activity above the photosphere. (Select all that apply)

A. Prominences are huge loops of the Sun’s ionized but cool material that are pushed by magnetic forces from the chromosphere into the corona.

B. There are regions of higher density and temperature than the surrounding material in the chromosphere called plages.

C. Flares are long lived phenomena above the photosphere beginning at the solar minimum and ending at solar maximum.

D. Coronal mass ejections happen when a flare is so violent that the material exceeds the escape speed of the sun. It is connected to sunspots.

Answers

There are up to four different kinds of features that can be seen when viewing the photosphere. Sunspots, faculae, granulation, and super-granulation are listed in decreasing order of observational ease. All the given statement are true.

Solar activityIn the chromosphere, plateaus are areas that are more dense and hotter than the surrounding material. Large, chilly, ionized loops of material known as prominences are slowly pushed into the corona by magnetic energy from the chromosphere. Flares are erratic, fiery, and intensely energetic bursts that last only a moment or two. When a flare is extremely powerful, its material is thrown into the solar system beyond the Sun's escape velocity and causes coronal mass ejections. These are all essentially connected to sunspots.

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A 0.675 mole sample of oxygen gas has a pressure of 2.50 atm at 50. °C. What volume of gas is present
in the sample? Show the rearranged ideal gas law solving for volume. Cancel units in work.

Answers

Considering the ideal gas law, the volume of gas present in the sample is 7.15122 L.

Definition of Ideal Gas Law

Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar gas constant:

P×V = n×R×T

Volume of gas in this case

In this case, you know:

P= 2.50 atmV= ?n= 0.675 molesR= 0.082 (atmL)/(molK)T= 50 C= 323 K (being 0 C= 273 K)

Replacing in the ideal gas law:

2.50 atm×V = 0.675 moles×0.082 (atmL)/(molK)× 323 K

Solving:

V= (0.675 moles×0.082 (atmL)/(molK)× 323 K)÷ 2.50 atm

V= 7.15122 L

Finally, the volume of gas is 7.15122 L.

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Which of the following reactants lead to the fastest reaction?
A) Solid aluminum and oxygen
B) Sodium hydroxide and hydrogen chloride in water
C) Copper nitrate and ethanol in water
D) Solid barium chloride and sodium sulfate​

Answers

Solid aluminum and oxygen are the reactant that lead to the fastest reaction. Therefore, the correct option is option A.

Chemical reaction, the transformation of one or more chemicals (the reactants) into one or more distinct compounds (the products). Chemical elements or chemical compounds make up substances.

Chemical reactions constitute a fundamental component of life itself, as well as technology and culture. Burning fuels, melting iron, creating glass or pottery, brewing beer, making wine. Solid aluminum and oxygen are the reactant that lead to the fastest reaction.

Therefore, the correct option is option A.

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Under what conditions would one mole of methane gas, ch4 , occupy the smallest volume?

Answers

Under the condition A. 273 K and 1.01 × [tex]10^{5}[/tex] Pa, one mole of methane gas, [tex]CH_{4}[/tex], occupy the smallest volume.

What is the smallest volume ?

The ideal gas law, PV = nRT, can be rearranged to solve for volume as V = nRT/P, where V is volume, n is the number of moles, R is the gas constant, T is temperature in Kelvin, and P is pressure.

To find the conditions under which one mole of methane gas occupies the smallest volume, we need to find the combination of temperature and pressure that results in the smallest volume.

Using the equation V = nRT/P, we can see that volume is inversely proportional to pressure, so we want the highest pressure possible. Therefore, we can eliminate choices A and C, which have lower pressures than B and D.

Next, we need to consider temperature. Using the same equation, we can see that volume is directly proportional to temperature, so we want the lowest temperature possible. Therefore, the correct choice is A, which has a temperature of 273 K, the lowest temperature of all the choices.

Therefore, the answer is A. 273 K and 1.01  × [tex]10^{5}[/tex]  Pa.

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Complete question is: Under the 273 K and 1.01 × [tex]10^{5}[/tex] Pa condition, one mole of methane gas, [tex]CH_{4}[/tex], occupy the smallest volume.

A sample of helium gas, He(g), is placed in a rigid cylinder sealed with a movable piston.
The temperature of the helium is 25.0°C. The volume of the helium is 300. milliliters and
the pressure is 0.500 atmosphere.

16) State, in terms of the average distance between the helium atoms, why the density of the gas increases when the piston is pushed farther into the rigid cylinder. [1]

17) Determine the volume of the helium gas when the pressure is increased to 1.50 atm and
the temperature remains at 25.0°C. [1]

Answers

(16) When the piston is pushed farther into the rigid cylinder, the volume available for the helium gas decreases. This leads to an increase in density because the gas molecules are packed into a smaller volume, resulting in a shorter average distance between the helium atoms; (17) volume = 100mL.

What will be the effect on density on pushing the piston  farther into the rigid cylinder?

When the piston is pushed farther into the rigid cylinder, the volume available for the helium gas decreases. This causes the average distance between the helium atoms to decrease as well. Since the volume available for the gas molecules to move in is reduced, they will collide more frequently with each other and with the walls of the container, leading to an increase in density.

(17) To determine the new volume of the helium gas, we can use the combined gas law:

(P1V1)/T1 = (P2V2)/T2

where P1, V1, and T1 are the initial pressure, volume, and temperature of the helium gas, and P2 and T2 are the final pressure and temperature.

Substituting the given values, we get:

(0.500 atm)(300. mL)/(298.15 K) = (1.50 atm)(V2)/(298.15 K)

Solving for V2, we get:

V2 = (0.500 atm)(300. mL)/(1.50 atm) = 100. mL

Therefore, the volume of the helium gas when the pressure is increased to 1.50 atm and the temperature remains at 25.0°C is 100mL.

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Problem 1. Sea water contains dissolved salts at a total ionic concentration of about 1.13 mol×L–1. What pressure must be applied to prevent osmotic flow of pure water into sea water through a membrane permeable only to water molecules (at 25oC)?
Problem 2. What is the osmotic pressure of a solution prepared by adding 6.65 g of glucose to enough water to make 350 mL of solution at 35°C?
Problem 3. What is the osmotic pressure of a solution prepared by adding 9.0 g of glucose to enough water to make 450 mL of solution at 35°C?
Problem 4. What is the osmotic pressure of a solution prepared by adding 11.0 g of propanol to enough water to make 850 mL of solution at 25°C?
Problem 5. What is the osmotic pressure of a solution prepared by adding 65 g of glucose to enough water to make 35000 mL of solution at 15°C?

Answers

Problem 1:

The osmotic pressure can be calculated using the equation:

π = CRT

Where π is the osmotic pressure, C is the total concentration of dissolved particles, R is the gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.

In this case, the total concentration of dissolved particles is 1.13 mol/L, and the temperature is 25°C, or 298 K.

π = (1.13 mol/L) (0.0821 L·atm/(mol·K)) (298 K) = 24.7 atm

Therefore, a pressure of 24.7 atm must be applied to prevent osmotic flow of pure water into seawater.

Problem 2:

The osmotic pressure of a solution can be calculated using the equation:

π = MRT

Where M is the molar concentration of the solution, R is the gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.

First, we need to calculate the molar concentration of the glucose solution:

molar mass of glucose = 180.16 g/mol
moles of glucose = 6.65 g / 180.16 g/mol = 0.0369 mol
volume of solution = 350 mL = 0.350 L

Molar concentration = moles of solute / volume of solution = 0.0369 mol / 0.350 L = 0.105 M

Now we can calculate the osmotic pressure:

π = (0.105 M) (0.0821 L·atm/(mol·K)) (308 K) = 2.87 atm

Therefore, the osmotic pressure of the glucose solution is 2.87 atm.

Problem 3:

Using the same equation as in Problem 2:

molar mass of glucose = 180.16 g/mol
moles of glucose = 9.0 g / 180.16 g/mol = 0.0499 mol
volume of solution = 450 mL = 0.450 L

Molar concentration = moles of solute / volume of solution = 0.0499 mol / 0.450 L = 0.111 M

π = (0.111 M) (0.0821 L·atm/(mol·K)) (308 K) = 3.05 atm

Therefore, the osmotic pressure of the glucose solution is 3.05 atm.

Problem 4:

In this case, propanol is the solute, so we need to first calculate the molar concentration:

molar mass of propanol = 60.10 g/mol
moles of propanol = 11.0 g / 60.10 g/mol = 0.183 mol
volume of solution = 850 mL = 0.850 L

Molar concentration = moles of solute / volume of solution = 0.183 mol / 0.850 L = 0.215 M

π = (0.215 M) (0.0821 L·atm/(mol·K)) (298 K) = 4.84 atm

Therefore, the osmotic pressure of the propanol solution is 4.84 atm.

Problem 5:

molar mass of glucose = 180.16 g/mol
moles of glucose = 65 g / 180.16 g/mol = 0.361 mol
volume of solution = 35000 mL = 35.00 L

Molar concentration = moles of solute / volume of solution = 0.

Problem 1:

The osmotic pressure (π) can be calculated using the van 't Hoff equation: π = iMRT, where i is the van 't Hoff factor (1 for water), M is the molar concentration, R is the gas constant, and T is the temperature in Kelvin.

In this case, the molar concentration is 1.13 mol/L, and the temperature is 25°C = 298 K. So,

π = iMRT = (1)(1.13 mol/L)(0.08206 L·atm·K⁻¹·mol⁻¹)(298 K)

π = 29.8 atm

Therefore, a pressure of 29.8 atm must be applied to prevent osmotic flow of pure water into sea water through a membrane permeable only to water molecules at 25°C.

Problem 2:

The osmotic pressure of a solution can be calculated using the van 't Hoff equation: π = iMRT, where i is the van 't Hoff factor, M is the molar concentration, R is the gas constant, and T is the temperature in Kelvin.

First, we need to find the molar concentration of glucose in the solution. The molecular weight of glucose is 180.16 g/mol. So,

Molar concentration = (mass/volume) / (molecular weight)

Molar concentration = (6.65 g/0.35 L) / 180.16 g/mol

Molar concentration = 0.104 mol/L

Now, we can calculate the osmotic pressure:

π = iMRT = (1)(0.104 mol/L)(0.08206 L·atm·K⁻¹·mol⁻¹)(308 K)

π = 2.44 atm

Therefore, the osmotic pressure of the solution prepared by adding 6.65 g of glucose to enough water to make 350 mL of solution at 35°C is 2.44 atm.

Problem 3:

Using the same process as in Problem 2, we can find the molar concentration of glucose in the solution:

Molar concentration = (mass/volume) / (molecular weight)

Molar concentration = (9.0 g/0.45 L) / 180.16 g/mol

Molar concentration = 0.44 mol/L

Now, we can calculate the osmotic pressure:

π = iMRT = (1)(0.44 mol/L)(0.08206 L·atm·K⁻¹·mol⁻¹)(308 K)

π = 10.2 atm

Therefore, the osmotic pressure of the solution prepared by adding 9.0 g of glucose to enough water to make 450 mL of solution at 35°C is 10.2 atm.

Problem 4:

Propanol (C₃H₇OH) is a non-electrolyte, so its van 't Hoff factor is 1.

First, we need to find the molar concentration of propanol in the solution. The molecular weight of propanol is 60.10 g/mol. So,

Molar concentration = (mass/volume) / (molecular weight)

Molar concentration = (11.0 g/0.85 L) / 60.10 g/mol

Molar concentration = 0.178 mol/L

Now, we can calculate the osmotic pressure:

π = iMRT = (1)(0.178 mol/L)(0.08206 L·atm·K⁻¹·mol⁻¹(298 K)

π = 3.67 atm

Therefore, the osmotic pressure of the solution prepared by adding 11.0 g of propanol to enough water to make 850 mL of solution at 25°C is 3.67 atm.

Problem 5:

Using the same process as in Problem 2 and Problem 3, we can find the molar concentration of glucose in the solution:

Molar concentration = (mass/volume) / (molecular weight)

Molar concentration = (65 g/35,000 mL) / 180.16 g/mol

Molar concentration = 0.00177 mol/L

Now, we can calculate the osmotic pressure:

π = iMRT = (1)(0.00177 mol/L)(0.08206 L·atm·K⁻¹·mol⁻¹)(288 K)

π = 0.0398 atm

Therefore, the osmotic pressure of the solution prepared by adding 65 g of glucose to enough water to make 35,000 mL of solution at 15°C is 0.0398 atm.

How many carbon atoms are in 11.5 g C2H5OH?

Answers

There are approximately 3.02 x 10²³ carbon atoms in 11.5 g of [tex]C_{2}H_{5} OH[/tex]

To calculate the number of carbon atoms present in the 11.5g of [tex]C_{2}H_{5}OH[/tex]which is ethanol, we need to find and calculate the number of moles of ethanol present and convert moles to the number of carbon atoms by using Avogadro's number.

The molecular weight of [tex]C_{2}H_{5}OH[/tex] is =

no. of atoms x its atomic weight. So the molecular weight is,

(2 atoms of carbon x 12 g/mol )+ (5 atoms of hydrogen x 1 g/mol) + (1 atom of oxygen x 16 g/mol) + (1 atom of hydrogen x 1 g/mol ) =

24+10+16+1  ≅  47 g/mol

The number of moles of ethanol = given mass / molecular weight = 11.5g /47  g/mol =  0.2446 moles

Number of Carbon atoms = no. of moles of ethanol x Avogadro number x no. of carbon atoms in each molecule of ethanol =  0.2446 moles x 6.022 x 10²³ atoms/mol x 2 carbon atoms/molecule ≅ 3.02 x 10²³

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Sucrose (C12H22011) is combusted in air according to the following reaction:
C12H22011(s) + O2(g) = CO2(g) + H2O(l )
How many moles of carbon dioxide would be produced by the complete combustion of 38.5 grams of sucrose in the presence of excess oxygen?

Answers

Okay, let's break this down step-by-step:

1) The molecular formula for sucrose is C12H22011. This means each mole of sucrose contains 12 moles of carbon and 22 moles of hydrogen.

2) You are combusting 38.5 grams of sucrose. To convert grams to moles, we divide by the molar mass:

Molar mass of C12H22011 = 342.3 g/mol

So 38.5 g / 342.3 g/mol = 0.113 moles of sucrose

3) According to the combustion reaction, each mole of sucrose produces 12 moles of CO2.

So 0.113 moles of sucrose will produce 0.113 * 12 = 1.356 moles of CO2.

4) Round to the nearest whole number:

1 mole of CO2

Therefore, the complete combustion of 38.5 grams of sucrose in excess oxygen would produce 1 mole of carbon dioxide.

Let me know if you have any other questions!

What is the temperature of 0.59 mol of gas at a pressure of 1.0 atm and a volume of 11.0 L? (Express the temperature in kelvins to two significant figures).

Answers

Answer:

T = 230 K

Explanation:

Simply use the equation: PV = nRT, where P is pressure, V is volume, n is moles, R is 0.0821, and T is temperature in kelvin

Place the values in

(1.0 atm)(11.0 L)=(0.59 moles)(0.0821 Lxatm/Kxmol)(T)

Solve for T = 230 K

What is the energy change for the following equation?

Answers

Answer:

1) -572 kJ/mol

2) -2220.7 kJ/mol

Explanation:

Multiply the bond enthalpies to take into account the amount of moles of each compound in the reaction. Then, to get the total change in energy/enthalpy in the reaction, subtract the reactant energy from the product energy.

1) 2(-286) = -572 kJ/mol

2) 4(-286) + -393.5(3) - -103.8 = -2220.7 kJ/mol

For which of the following reactions does ΔH
o
rxn
= ΔH
o
f
?

(a) H2(g) + S(rhombic) → H2S(g)

(b) C(diamond) + O2(g) → CO2(g)

(c) H2(g) + CuO(s) → H2O(l) + Cu(s)

(d) O(g) + O2(g) → O3(g)

Answers

ΔH° = ΔHf° is true for two reactions which are ,

H₂(g) + S(rhombic) → H₂S(g)

C(diamond) + O₂(g) → CO₂(g)

Hence, option a and b are correct.

∆Hf is the enthalpy of formation, is the enthalpy when product is made from its constituent elements. Generally the enthalpy of formation is described as the standard reaction enthalpy for the formation of the compound from its elements (which may include atoms or molecules) in their most stable reference states at the chosen temperature (298.15K) and at 1bar pressure.

In a part, H₂S is formed by H₂ and S means it's constituent elements.

So, a is correct.

In b also, CO₂ is formed by C and O₂, so b is also correct. Hence, option a and b are correct.

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Consider a sample containing 0.300 mol of a substance. How many atoms are in the sample if the substance is iron?

Answers

The number of atoms in a sample containing 0.300 mol of iron is 1.8066 x 10²³ atoms.

The number of atoms in a sample of a substance can be determined using Avogadro's number, which is equal to 6.022 x 10²³ particles per mole. The atomic mass of iron is 55.85 g/mol.

Therefore, 0.300 mol of iron will contain:

0.300 mol x 6.022 x 10³ atoms/mol = 1.8066 x 10²³ atoms

Avogadro's number is a fundamental constant in chemistry that represents the number of particles (atoms or molecules) in one mole of a substance.

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What is molecular weight of a substance given that 1.22g of the sample was vaporised in 100ml flask at 45°C and 687mmHg.​

Answers

To calculate the molecular weight of a substance, we can use the ideal gas law, which relates the pressure (P), volume (V), number of moles (n), and temperature (T) of a gas:

PV = nRT

where R is the gas constant. Rearranging this equation gives:

n = PV / RT

We can use this equation to calculate the number of moles of the substance in the vapor phase. Then, we can use the definition of molecular weight, which is the mass of one mole of the substance, to calculate the molecular weight.

First, let's convert the temperature and pressure to SI units:

Temperature: 45°C = 318.15 K
Pressure: 687 mmHg = 91.6 kPa

Next, let's calculate the number of moles of the substance in the vapor phase. We know that 1.22 g of the substance was vaporized in a 100 mL flask, which is equivalent to 0.1 L. Therefore, the concentration of the substance in the vapor phase is:

c = 1.22 g / 0.1 L = 12.2 g/L

We can convert this concentration to units of pressure using the ideal gas law:

PV = nRT

n/V = P/RT

n/V = (91.6 kPa) / (8.314 J/mol K * 318.15 K) = 0.0361 mol/L

Therefore, the number of moles of the substance in the vapor phase is:

n = (0.0361 mol/L) * 0.1 L = 0.00361 mol

Finally, we can calculate the molecular weight using the definition of molecular weight:

molecular weight = mass / moles

mass = 1.22 g

moles = 0.00361 mol

molecular weight = 1.22 g / 0.00361 mol = 337.4 g/mol

Therefore, the molecular weight of the substance is approximately 337.4 g/mol.

Classify each of the following aqueous solutions as a nonelectrolyte, weak electrolyte, or strong electrolyte.
(a) PbCl2
(b) N(CH3)3
(c) CsOH
(d) H2S
(e) CrCl2
(f) Ni(CH3COO)2

Answers

Strong electrolyte: lead(II) chloride weak electrolyte trimethylamine Strong electrolyte: cesium hydroxide sulphide of hydrogen: a weak electrolyte, strong electrolyte chromium chloride Weak electrolyte: nickel(II) acetate

Is acetic acid an electrolyte that is weak?

Response and justification The fact that acetic acid has a low dissociation constant suggests that it is a weak acid. Acetic acid's restricted ability to ionise in an aqueous solution is a result of its low dissociation constant. With this in mind, acetic acid can be regarded as a weak electrolyte.

Is acid phosphoric a type of electrolyte?

Phosphoric acid is a useful electrolyte because of its low volatility, strong ionic conductivity, stability at high temperatures, carbon dioxide and carbon monoxide tolerance, and low flammability.

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Calculate the molality of a solution prepared by dissolving 19.9 g of KCl in 750.0 mL of water.

Answers

Answer:

Explanation:

First, we need to calculate the moles of KCl:

Calculate the molar mass of KCl:

KCl = 39.10 g/mol (atomic weight of K) + 35.45 g/mol (atomic weight of Cl)

= 74.55 g/mol

Calculate the moles of KCl:

moles of KCl = mass of KCl / molar mass of KCl

= 19.9 g / 74.55 g/mol

= 0.267 mol

Next, we need to convert the mass of the solvent (water) from milliliters to kilograms:

Convert mL to L:

750.0 mL = 750.0 mL * (1 L / 1000 mL)

= 0.7500 L

Calculate the mass of water:

mass of water = volume of water x density of water

= 0.7500 L x 1000 g/L

= 750.0 g

Convert the mass of water to kilograms:

mass of water = 750.0 g / 1000 g/kg

= 0.7500 kg

Now we can calculate the molality of the solution:

molality = moles of solute / mass of solvent (in kg)

molality = 0.267 mol / 0.7500 kg

= 0.356 mol/kg

Therefore, the molality of the solution is 0.356 mol/kg.

Macmillan Learning
A 346.9 mL sample of carbon dioxide was heated to 373 K. If the volume of the carbon dioxide sample at 373 K is 596.2 mL,
what was its temperature at 346.9 mL?
T=. K

Answers

Answer:

Explanation:

Use the Equation: [tex]\frac{V1}{T1}=\frac{V2}{T1} \\[/tex]

V1 = 346.9 mL

T1 = ? K

V2 = 596.2 mL

T2 = 373 L

This assumption can only be made when pressure is held constant.

[tex]\frac{346.9}{V1}=\frac{596.2}{373}, solve for V1\\ V1 = 217.0 K[/tex]

Determine which statements are consistent with Daltons atomic theory

Answers

Answer:

Explanation:

Dalton's atomic theory had several postulates, including:

1. All matter is made up of atoms, which are indivisible and indestructible.

2. All atoms of a given element are identical in mass and properties.

3. Compounds are formed by a combination of two or more different kinds of atoms.

4. A chemical reaction is a rearrangement of atoms.

Based on these postulates, the following statements are consistent with Dalton's atomic theory:

1. Chemical reactions involve the rearrangement of atoms to form new compounds.

2. Elements are composed of small, indivisible particles called atoms.

3. Atoms of the same element have the same mass and chemical properties.

4. A compound is made up of two or more different elements.

5. Atoms can neither be created nor destroyed in a chemical reaction.

However, the following statements are not consistent with Dalton's atomic theory:

1. Atoms are divisible into smaller particles.

2. Isotopes of the same element have different chemical properties.

3. The mass of an atom is distributed uniformly throughout the atom.

It's worth noting that while Dalton's atomic theory has been largely superseded by modern atomic theory, it remains an important historical milestone in the development of atomic theory.

please help this is dus today please help ​

Answers

A percentage expressing the variation between a calculated/measured value and the anticipated/real value is known as a "calculated percent error."

What does this mean?

A positive numerical discrepancy implies that the assessed number exceeds what was expected, with negative values indicating the reverse.

When conducting an experiment, if an excessive percent error emerges it suggests potential errors or miscalculations in experimental design, data collection, or analysis.

This may mean unaccounted variables were present, and equipment/procedures require refinement. In opposition, lower percentages signal accuracy despite any limitations resulting from the human elements involved.

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A sample of sodium azide (NaN3), a compound used in automobile air bags, was thermally decomposed, and 15.3 mL nitrogen gas was collected over water at 25°C and 755 torr. Given the vapour pressure of water at 25°C is 23.6 torr, how many grams of nitrogen were collected? ​

Answers

A sample of sodium azide (NaN3), a compound used in automobile air bags, was thermally decomposed, and 15.3 mL nitrogen gas was collected over water at 25°C and 755 torr. 131.1g is the mass of nitrogen.

In physics, mass is a quantitative measurement of inertia, a basic characteristic of all matter. It essentially refers to a body of matter's resistance to changing its speed or location in response to the application for a force.

The change caused by a force being applied is smaller the more mass a body has. The kilogramme, which is defined approximately equal to 6.62607015 1034 joule second in terms of Planck's constant, is the unit of mass within the Internacional System of Units (SI).

P×V = n×R×T  

755×15.3 = n×0.0821×300

11551.5=n×24.63

n= 46.9

mass =  46.9×28=131.1g

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Calculation: 3. A new alloy of steel is 525 g at 100-C. It is dropped into 375 grams of water at 25 C. The final temperature changes to 55C, what is the specific heat of steel? ​

Answers

The specific heat capacity of the steel is 2 g/°C.

What is the specific heat capacity of the steel?

To determine the specific heat capacity of a particular sample of steel, the sample can be heated to a known temperature and then placed in contact with a known amount of water at a lower temperature.

We know that the heat lost by the steel is equal to the heat gained by the water.

Thus we have that;

-(525 * c * (55- 100)) = (375 * 4.2 * (55- 25)

23625c = 47250

c = 47250/23625

c = 2 g/°C

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How many chloride ions are in 15.0 mL of a 2.5 molar solution of magnesium chloride? (MgCl2)

Answers

In 15.0 mL of a 2.5 molar magnesium chloride solution, there are approximately 4.51 × 10^{22} chloride ions.

How many chloride ions are there in MgCl_2?

An inorganic compound composed of one magnesium ion and two chloride ions.

We must first determine the number of moles of MgCl_2 in the solution,

moles of solute = concentration × volume (in liters)

Converting the solution's volume from milliliters to liters,

15.0 mL = 15.0 × 10^{-3} L

moles of MgCl_2 = 2.5 mol/L × 15.0 × 10^{-3} L = 0.0375 moles

The solution contains the following number of chloride ions:

number of Cl^{-} ions = 2 × moles of MgCl_2

Substitute the value of moles of MgCl_2,

number of Cl^{-} ions = 2 × 0.0375 moles = 0.075 moles

by using Avogadro's number:

number of Cl^{-} ions = 0.075 moles × 6.02 × 10^{23} ions/mol ≈ 4.51 × 10^{22} ions

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3. A solution of hydrochloric acid is made by dissolving hydrogen chloride gas in 100.0ml water. This solution neutralizes a 15ml sample of 0.10 mol/L sodium carbonate solution. a. What mass of hydrogen chloride gas was dissolved in 100.0ml of water? b. What volume of hydrogen chloride was this?​

Answers

Okay, here are the steps to solve this problem:

a) To neutralize 15ml of 0.10 mol/L sodium carbonate solution, the hydrochloric acid solution must contain 0.015 moles of HCl.

Since the HCl solution is made by dissolving HCl gas in 100ml water, we can calculate the moles of HCl gas dissolved in 100ml:

0.015 moles HCl / 15ml sodium carbonate solution = X moles HCl gas / 100ml HCl solution

X = 0.015 * (100/15) = 0.01 moles HCl gas

b) Molar volume of HCl gas at STP is 24.45 L/mol.

So the volume of 0.01 moles HCl gas is: 0.01 moles * 24.45 L/mol = 0.2445 L

Since the solution is made with this gas in 100ml water, the volume of HCl gas dissolved in 100ml water is 0.2445 L.

So the final answers are:

a) 0.01 moles of HCl gas

b) 0.2445 L of HCl gas

Please let me know if you have any other questions!

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