The quantity of antimony in a sample can be determined by an oxidation–reduction titration with an oxidizing agent. A 7.97 g sample of stibnite, an ore of antimony, is dissolved in hot, concentrated HCl(aq) and passed over a reducing agent so that all the antimony is in the form Sb3+(aq). The Sb3+(aq) is completely oxidized by 34.8 mL of a 0.130 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is

BrO−3(aq)+Sb3+(aq)⟶Br−(aq)+Sb5+(aq)(unbalanced)

Calculate the amount of antimony in the sample and its percentage in the ore.


mass of antimony in grams:

percentage of antimony:

Answer: No.

Explanation:

The freezing point if 30% ethylene glycol is around -14.7°C, so the plane cannot take off even if it was de-iced. Jet fuel has a very low freezing temperature of around -40°C, but the fuel cannot be the only thing unfrozen for the plane to continue on successfully. If you alter the ethylene glycol concentration to 60%, however, the freezing point drops to -60°C, which would be perfect for this scenario. This happens because the hydrogen bonds that have to form between the water molecules to freeze are interrupted by ethylene glycol molecules, and vice versa.

0.0001212598 is a long will it be for one-half of carbon-14 in a sample to remain.

A reaction's half-life is the amount of time needed for the reactant concentration to drop to half its initial value. The rate constant for a first-order reaction and its half-life are connected by the constant t1/2 = 0.693/k.

The half-life of carbon-14 is 5,730 40 years, meaning that during the following 5,730 years, half of the radioisotope that is present at any particular time will spontaneously disintegrate.

Carbon-14's 5715-year half-life

The decay constant needs to be determined.

We know the half-life equation.

t[tex]\frac{1}{2}[/tex] = 0.693/λ

where,

t1/2  ​is the half-life and λ is the decay constant

so,

λ = 0.693 / 5715 = 0.0001212598

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Answer:

algebraic reaction is the answer of this question

The percent yield for the reaction below when 364 g SO₂ and 42.0 g O₂ produce 408 g SO₃ is 51.5 %.

The reaction is given as :

2SO₂(g) + O₂(g)   ------>   2SO₃

mass of SO₂ = 364 g

mass of O₂ = 42 g

mass of SO₃ = 408 g

moles of SO₂= mass / molar mass

                     = 364 / 64

                     = 5.68 mol

moles of O₂ = mass / molar mass

                    = 42 / 32

                   = 1.31 mol

1 mole of O₂ react with = 2 mole of  SO₂

1.31 mole of O₂ react with = 2 × 1.31 = 2.62 mol of  SO₂

this means O₂ is limiting reagent.

1 mole of O₂  produce 2 mole of SO₃

1.31 mole of O₂  = 2 × 1.31 = 2.62 mol of SO₃

mass of SO₃ = 2.62 × 80

                     = 209.6 g

Percent yield =( actual yield / theoretical yield ) × 100 %

                      = (209.6 / 408) × 100 %

                      = 51.5 %

Thus, The percent yield for the reaction below when 364 g SO₂ and 42.0 g O₂ produce 408 g SO₃ is 51.5 %.

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The force of gravity needed to make a 13kg ball fall on Earth is 127.53N.

Force of gravity is a force that attracts any two objects with mass. The force of gravity is the weight of an object on Earth.

The weight of the force on an object due to the gravitational attraction between it and the Earth (or whatever astronomical object it is primarily influenced by).

The weight of an object can be calculated by multiplying the force of the object by acceleration due to gravity on Earth as follows:

Weight = force × 9.81m/s²

Weight = 13kg × 9.81m/s²

Weight = 127.53N.

Therefore, 127.53N is the weight of the ball on Earth.

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Answer:

1.67262192 × 10-27 kilograms

The mass of a proton is 1.6726219 × 10-27 kilograms.

What do protons do?

Proton properties:

Proton Mass: One proton has the same mass as a hydrogen atom. A proton and a electron together make up a hydrogen atom. The mass of a hydrogen atom is equal to the mass of a proton because the mass of an electron is thought to be negligible. An electron weighs 1840 times as much as a proton.

Proton mass is equal to 1.676 x 1027 kilograms, or 1.676 x 1024 grams.

Charge of Proton - A proton's charge is equal to and in the opposite direction of an electron's. Because of this, it has a unit positive charge.

A proton has a charge of +1.602 10-19 coulombs.

Hence, The mass of a proton is 1.6726219 × 10-27 kilograms.

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The number of atoms of phosphorus that would be in 4.70 mol of copper (II) phosphate is 56.61 x [tex]10 ^{23[/tex] atoms.

According to an established rule, 1 mole of any substance contains one Avogadro's number, which is 6.022 x [tex]10 ^{23[/tex] atoms.

Copper (II) phosphate contains atoms of copper, phosphorus, and oxygen. In ion form, the compound ionizes according to the following equation:

[tex]Cu_3(PO_4)_2 --- > 3 Cu^2^+ + 2P^4^+ + 8O^2^-[/tex]

This means that every 1 mole of copper (II) phosphate contains 2 moles of phosphorus. The equivalent mole of phosphorus in 4.70 mol of copper (II) phosphate would be:

= 4.70 x 2

= 9.4 mol of phosphorus.

We said 1 mole of any substance =  6.022 x [tex]10 ^{23[/tex] atoms.

Hence, 9.4 moles of phosphorus will contain:

9.4 x  6.022 x [tex]10 ^{23[/tex] atoms

= 56.61 x [tex]10 ^{23[/tex] atoms

In other words, 4.70 mol of copper (II) phosphate will contain  56.61 x [tex]10 ^{23[/tex] atoms of phosphorus.

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A+B⇄ D (fast equilibrium) is  the rate of formation of C .

Equilibrium is a dynamic process often occurs while chemical reaction. All of the compounds have constant concentrations when the forward and reverse reaction speeds are equal. Given this, it makes reasonable that the concentration ratio for any given reaction will always maintain a constant value at equilibrium. The equilibrium constant is the proportion of a reaction's products' mathematical product to their reactant concentrations (Keq). Each concentration is increased to the power of its associated coefficient by the balanced chemical equation. When equilibrium is reached, K = 1.

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The relationship that exists between smooth endoplasmic reticulum and cell membrane is that lipids are produced in the smooth endoplasmic reticulum which is the major product used for the synthesis of cell membrane.

The cell membrane is defined as part of the cell that is semi permeable which surrounds the cells and controls the passage of substances in and out of the cell.

The endoplasmic reticulum is a membrane bound organelle that are of two types namely:

The smooth endoplasmic reticulum is defined as the type of endoplasmic reticulum that is involved in the synthesis of lipids such as cholesterol and phospholipids.

A relationship exists between the cell membrane and the smooth endoplasmic reticulum because the lipids required for the synthesis of cell membrane are being produced by the smooth endoplasmic reticulum.

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The redox reaction as it has been written can not be able to hold.

We know that a redox reaction is the kind of reaction that involves the oxidation and the reduction of a specie. In this case, we can see the reaction that occurs between the aluminum ion and sodium sulfide.

We know that that reaction can not be able to hold because the reduction potential of the aluminum ion is -1.66 V while the reduction potential of sodium is -2.71 V.

as such, sodium is more reducing than aluminum hence it can not be displaced by it.

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Answer:

Q = mcΔT

Q = (800. g)(4.186 J/g°C)(8.00°C)

Q = 26,800 J

Explanation:

To calculate the thermal energy of the water, you need to use the following equation:

Q = mcΔT

In this equation,

-----> Q = thermal energy (J)

-----> m = mass (g)

-----> c = specific heat capacity (J/g°C)

-----> ΔT = T₂ - T₁ = change in temperature (°C)

"T₁" represents the initial temperature and "T₂" represents the final temperature. The specific heat capacity of water is 4.186 J/g°C. You can plug the given values and the constant into the equation and solve for "Q" to find the thermal energy. The final answer should have 3 sig figs like the given values.

Q = ? J                              c = 4.186 J/g°C

m = 800. g                       ΔT = T₂ - T₁ = 20.5°C - 12.5°C = 8.00°C

Q = mcΔT                                                        <----- Given equation

Q = (800. g)(4.186 J/g°C)(8.00°C)                  <----- Insert values

Q = 26,800 J                                                  <----- Multiply

The numerical setup for calculating the change in the thermal energy of the water in the bottle from 7 a.m. to 3 p.m. is: ΔQ = (800 g) * (4.18 J/g·°C) * (8°C) = 26,528 J

To calculate the change in thermal energy of the water in the bottle from 7 a.m. to 3 p.m., you can use the formula:

ΔQ = mcΔT

where:

ΔQ is the change in thermal energy,

m is the mass of the water,

c is the specific heat capacity of water, and

ΔT is the change in temperature.

Given data:

Mass of water (m) = 800 g

Specific heat capacity of water (c) = 4.18 J/g·°C

Initial temperature (T1) at 7 a.m. = 12.5°C

Final temperature (T2) at 3 p.m. = 20.5°C

Now, plug in the values into the formula:

ΔQ = (800 g) * (4.18 J/g·°C) * (20.5°C - 12.5°C)

Simplifying the expression:

ΔQ = (800 g) * (4.18 J/g·°C) * (8°C)

ΔQ = 26,528 J

Therefore, the numerical setup for calculating the change in the thermal energy of the water in the bottle from 7 a.m. to 3 p.m. is:

ΔQ = (800 g) * (4.18 J/g·°C) * (8°C)

= 26,528 J

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The concentration in parts per million of DDT are as follow:

3 x 10^-4 g in 2.3 kg minnow tissue PPM DDT is 0.13 ppm and in 717 µg in 1.5 kg breast milk is 0.478 ppm.

ppm DDT

PPM is defined as parts per million. It represents the lower concentration of the given solution. It also help in representing 0.001 gram or a milligram in the 1000 mL, which is equivalent to the 1 mg per liter.

Given,

A sample size of2.3 kg contained 3 x 10^-1 mg DDT

It means, 2.3 L sample contained 3 x 10^-1 mg DDT

Therefore, we can say that,

in part per million

= 3 x 10^-1 mg/ 2.3 L

= 0.13 mg/L

=0.13 ppm

A sample size of 1.5 kg contained 717 µg DDT

It means, 1.5 L sample contained 717 x 10^-3 mg DDT

Therefore, we can say that,

in part per million

= 717 x 10^-3 mg/ 1.5 L

= 0.478 mg/L

=0.478 ppm

Thus, we concluded that the concentration in parts per million of DDT are as follow:

3 x 10^-4 g in 2.3 kg minnow tissue PPM DDT is 0.13 ppm and in 717 µg in 1.5 kg breast milk is 0.478 ppm.

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The volume in liters of 5.75 moles of CO₂ at standard temperature and pressure (STP) is 128.88L.

The volume of a given gas at STP can be calculated using the following formula;

PV = nRT

Where;

According to this question, 5.75 moles of CO₂ at standard temperature and pressure (STP). The volume of this gas can be calculated as follows;

At STP, the following applies;

1 × V = 5.75 × 0.0821 × 273

V = 128.88L

Therefore, 128.88L is the volume of the gas.

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The number of formula units that make up 26.4g of magnesium chloride is 1.67 × 10²³ units.

Formula units is the empirical formula of an ionic compound (that does not possess individual molecules) for use in stoichiometric calculations.

The number of formula units of a substance can be calculated by multiplying the number of moles of the substance by its number of moles as follows:

no of formula units = no of moles × 6.02 × 10²³

First, we convert the mass of magnesium chloride to moles by dividing by its molar mass. Molar mass of magnesium chloride is 95.211 g/mol.

moles = 26.4g ÷ 95.211g/mol = 0.277moles

no of formula units = 0.277moles × 6.02 × 10²³

no of formula units = 1.67 × 10²³ units

Therefore, 1.67 × 10²³ units is the number of formula units of magnesium chloride.

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The reaction is a double displacement reaction, a precipitate of silver chloride, AgCl, is obtained, and the unknown compound is barium chloride, BaCl₂

Double displacement reactions are reactions that occur when two soluble ionic compounds are mixed together, and a precipitate is formed as a result of the exchange of radicals between the metallic ions in the ionic compounds.

An example of a double displacement reaction is the reaction between barium chloride and silver nitrate, where a precipitate of silver chloride is obtained. The equation of the reaction is given below:

BaCl₂ (aq) + AgNO₃ (aq) ---> Ba(NO₃)₂ (aq) + AgCl₂ (s)

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The theoretical yield  of the precipitate is  1185 g and the balanced reaction equation is; [tex]3Pb(NO_{3} )_{2} (aq) + Ni_{2} (SO_{4}) _{3} (aq) -------- > 2Ni(NO_{3} )_{3}(aq) + 3PbSO_{4}(s)[/tex]

We know that the term chemical reaction equation has to do with the interaction that exists between reactants and product. Now we know that a balanced reaction equation ought to have the same number of atoms of each of the elements that were combined in the reaction.

We have the balanced reaction equation as;

[tex]3Pb(NO_{3} )_{2} (aq) + Ni_{2} (SO_{4}) _{3} (aq) -------- > 2Ni(NO_{3} )_{3}(aq) + 3PbSO_{4}(s)[/tex]

Now;

Number of moles of nickel III sulfate = 531 g/406 g/mol = 1.31 moles

Number of moles of lead II nitrate = 3.91 moles

Since 3 moles of nitrate reacts with 1 mole of the sulfate

x moles of the nitrate reacts with 1.31 moles of the sulfate

x = 3.93 moles

Thus  lead II nitrate is the limiting reactant.

If 3 moles of the nitrate produces 3 moles of the precipitate then

Theoretical yield = 3.91 moles  * 303 g/mol

= 1185 g of precipitate

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The concentration of the solution is obtained as 7.3 * 10^-6 M.

We know that one of the ways by which we are able to obtain the concentration of substances especially those that are colored is by the use of spectrophotometry. In a spectrophotometer, there is a wavelength of maximum absorption that could be used to measure the concentration of the solution as we can see in the question.

By the use of the Beer Lambert's law we have;

A = εcl

A = Absorption

ε = molar absorptivity

c = concentration

l = path length

Thus we would have;

0.2 = 27400 M-1cm-1 * c * 1.00 cm

c = 0.2/27400 M-1cm-1 *  1.00 cm

c = 7.3 * 10^-6 M

We would have from the calculation that we have a solution whose concentration is 7.3 * 10^-6 M.

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The molecular formula is (CH2O)5 when the scale factor is 5 and the empirical formula is CH2O.

Considering the question given above, the following data were obtained:

Empirical formula = CH₂O

Molar mass of compound = 150 g/mol

Scaling factor (n) =?

Empirical formula × n = molar mass

[CH₂O]n = 150

[12 + (2×1) + 16]n = 150

[12 + 2 + 16]n = 150

30n = 150

Divide both sides by 30

n = 5

Therefore, the scaling factor is 5

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1.57 x 10²⁴ atoms of carbon to moles of carbon is 6.72 mole

Mole is the SI unit of amount of substance of a specified elementary entity

Here given data is

Atoms of carbon =  1.57 x 10²⁴

We have to calculate moles of carbon = ?

So one mole of carbon specifies 6.022×10²³ individual carbon atom that is mole of carbon

Then  1.57 × 10²⁴ atoms/6.022×10²³C×atom×mol⁻¹

Moles of carbon = 6.72 mole

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0.4g of Hydrogen will be produced when 4.6g of Sodium is used to provide Sodium Hydroxide.

The balanced equation for the formation of Sodium Hydroxide is given below:

2Na + 2H₂O ------> 2NaOH + H₂

It is shown that 2 moles of Na produce 2moles of Hydrogen. We know that the atomic mass of Sodium is 23g/mol. Then the number of moles present in 4.6g of Sodium would be,

Moles of Sodium = 4.6/23

Moles of Sodium = 0.2mol

Therefore,4.6g of Sodium has 0.2mol. Therefore, then 0.2 mol of sodium reacts to give out 0.2 mol of Hydrogen gas

The mass of hydrogen gas could be computed by the product of the mass of hydrogen gas and moles of hydrogen which is given by,

Mass of 2 atoms of hydrogen = 2x1x0.2

Mass of 2 atoms of hydrogen= 0.4g

Therefore, for every 0.2mol of Sodium, 0.1 mol of Hydrogen is produced which is equal to 0.4g of Hydrogen.

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The number of grams of RbOH present in the solution can be found by using the formula

[tex]c = \frac{m}{M \times v} \\ [/tex]

where

m is the mass in grams

M is the molar mass in g/mol

v is the volume in L or dm³

making m the subject we have

m = c × M × V

First of all we have to find the molar mass of RbOH

M(RbOH) = 85.5 + 16 + 1 = 102.5 g/mol

v = 31 ml = 0.031 dm³ or 0.031 L

c = 5.6 M

Substituting it into the formula we have

m = 5.6 × 102.5 × 0.031 = 17.79 g

We have the final answer as

Hope this helps you

Answer:

Kc = 0.101 or 1.01*10^-1

Explanation:

Set up ICE table
2NH3 -> 3H2 + N2
I: 2 moles, 0 moles, 0 moles

C: -x moles   +y moles   +z moles

E: 1.25 moles 0+y moles   0+z moles

We know that x is 0.75 moles, since 2-0.75 is 1.25

for every mole of NH3 consumed, 1.5 moles of H2 are produced, and 0.5 moles of N2 are produced. If 0.75 moles of NH3 are consumed, then 1.5 * 0.75 moles of H2 are produced, and 1.5*0.5 moles of N2 are produced

so, y equals 1.5 * 0.75, which is 3/2 * 3/4, or 3/8, or 0.375 moles of H2 produced

z equals 1.5 * 0.5, which is 0.75 moles of N2 produced

Kc equals [tex]\frac{[N_2][H_2]^3}{[NH_3]^2}[/tex]

N2 = 0.75, N2 = 0.375, and NH3 = 1.25
So, Kc equals
(0.375 * 0.75^3)/(1.25^2), which equals 0.10125

Round to 3 sig figs to get 0.101

The nomenclature name of NaNO is Sodium Nitrite or could also be said as Sodium Nitrate.

Sodium nitrite or sodium nitrate is a chemical compound with Sodium cation and nitrate anion. So the combination of these two ions gives us Sodium nitrite.

It is a white, ionic, solid crystalline compound that is used as a food preservative and additive.

Sodium nitrite and sodium nitrate are both inorganic compounds with the chemical formula NaNO₃.

They are both white, crystalline solids that are highly soluble in water. Sodium nitrite is the active ingredient in many curing agents used to preserve meats, and it is also used as a color fixative in smoked fish. Sodium nitrate is used as a fertilizer and as an oxidizing agent in pyrotechnics.

Therefore the nomenclature name of the chemical compound is Sodium Nitrite or Sodium nitrate depending upon the anion NO₂ or NO₃

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The plant that should be moved here to protect from malnourishment is plant 3. The correct option is C.

The visible light spectrum is the portion of the electromagnetic spectrum that can be seen by the human eye. Simply put, this range of wavelengths is known as visible light.

The visible spectrum, also known as the optical spectrum, is the portion of the electromagnetic spectrum whose radiation consists of photons capable of sensitizing a normal person's human eye. The corresponding visible light radiation range is identified.

That being said, plant 3 is situated in front of a green stained glass window.

Plants have chlorophyll pigment that reflects back green light, making green light less effective for photosynthesis and causing them to die as quickly as if they were kept in the dark.

Thus, the correct option is C.

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According to the reaction illustrated, the amount, in moles, of copper (II) nitrate that will be formed is 0.456 mole.

The equation of the chemical reaction when silver nitrate reacts with copper is expressed as follows:

[tex]2AgNO_3 + Cu --- > Cu(NO_3)_2 +2 Ag[/tex]

The copper was in excess. This means that silver nitrate is the limiting reactant.

And 0.912 moles of the silver nitrate were reacted.

The mole ratio of silver nitrate and copper (II) nitrate according to the balanced equation of the reaction is 2:1. In other words, for every 1 mole of silver nitrate that reacts, 0.5 moles of copper (II) nitrate is formed.

Now, if 0.912 moles of silver nitrate is involved, the equivalent amount of moles of copper (II) nitrate that will be formed can be calculated as follows:

2 moles silver nitrate = 1 mole copper (II) nitrate

0.912 moles silver nitrate = 0.912 x 1/2

                                          = 0.456 mole

In other words, according to the reaction given, if 0.912 moles silver nitrate is reacted with excess copper, 0.456 mole copper (II) nitrate will be formed.

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Work is done when a gas expands from an initial volume of 122 mL to a final volume of 205 mL against an external pressure of 818 mmHg is 67.894 J

Work is the measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement

Here given data is

Initial volume = 122 mL

Final volume = 205 mL

External pressure =  818 mmHg

We have to calculate work done = ?

We know that work done against external pressure = W = -PV

Initial volume - Final volume = 205 mL - 122 mL = 83ml

W = -  818 mmHg×83ml

W = 67.894 J

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Taking into account the reaction stoichiometry, 11.2 liters of hydrogen gas can form at STP when 0.500 L of 2.00M HCl solution reacts with excess magnesium.

In first place, the balanced reaction is:

Mg + 2 HCl → H₂ + MgCl₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

First of all, you know that react 0.500 L of 2.00M HCl. Taking into account the definition of molarity. It is possible to calculate the number of moles that react as follows:

moles of HCl= 0.500 L× 2 M

moles of HCl= 1 mole

To calculate the number of moles H₂ formed, the following rule of three can be applied: if by reaction stoichiometry 2 moles of HCl form 1 mole of H₂, 1 mole of HCl form how many moles of H₂?

moles of H₂= (1 mole of HCl× 1 mole of H₂)÷ 2 moles of HCl

moles of H₂= 0.5 moles

Finally, to calculate the liters of H₂ formed, you can apply the following rule of three: if by definition of STP conditions 1 mole of H₂ occupies a volume of 22.4 liters, 0.5 moles occupies how much volume?

volume= (0.5 moles× 22.4 L)÷ 1 mole

volume= 11.2 L

In summary, 11.2 liters of hydrogen gas can be formed at STP.

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The concentration of the solution originally was 1.8 M.

We know that the concentration of the solution is obtained from the formula of the neutralization reaction which is;

CAVA/CBVB = NA/NB

CA =  concentration of the acid

CB = concentration of the base

VA = volume of acid

VB = volume of base

NA = number of moles of acid

NB = number of moles of base

Then

CAVANB = CBVBNA

CA = CBVBNA/VANB

CA =  0.1  * 25 * 1/14.2 * 1

= 0.18 M

Using the formula;

C1V1 = C2V2

25 * C1 =  0.18 * 250

C1 = 0.18 * 250/25

C1 = 1.8 M

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