Answer:approximately 5.627 × 1019 milliliters
Explanation:
(100 POINTS!!!!) In one to two sentences, describe the process by which the ionic compound, LiCl, would dissolve in the polar solvent, CH3COCH3
Answer:
LICL as an Ionic compounds has covalent ability and they are soluble in polar solvent such as CH3COCH3 but they are insoluble in non-polar solvents. Due to their polarity, CH3COCH3 will decreases the electrostatic forces of attraction thereby resulting in free ions in aqueous solution.
Water is the known as a polar solvent that can dissolve an ionic compounds very easy and as a polar solvent, the arrangement of oxygen and hydrogen atoms in water is in bent shape.
Ionic compounds like LICL that is very polar is soluble in the polar solvent water.
Polar solvents like CH3COCH3 will dissolve polar and ionic solutes because of the attraction of the opposite charges on the solvent and solute particles.
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Explanation:
Balancing Chemical Equations
Based on the chemical equation, use the drop down menu to choose the coefficients that will balance the chemical
equation
(vbo,- (v]),
Will make Brainlyest plzzzz help!!!
Answer:
3O2→ 2O3
Explanation:
by multiplying 2 by 3 we get 6 on both sides
Help with this please :) will mark Brainlyist if right
Answer:
3
Explanation:
What would be the atomic number of this atom?
Answer:
1 would be the answer
protons = atomic # I think
Answer:
hydrogen
Explanation:
hydrogen has the atomic number of 1 which means that hydrogen has one proton and is neutral so it has one electron as well.
Identify the major ionic species present in an aqueous solution of C6H12O6 (glucose).
A. 6 C-
, 12 H+
, 6 O–
B. 6 C+
, 12 H+
, 6 O2–
C. 6 CH2+, 6 O2–
D. C6
+
, 12 H+
, 6 O2–
E. no ions are present
Glucose is a molecular substance therefore, there are no ionic species in glucose.
Ionic substances dissolve in water to yield ions. Molecular substances do not produce ions in solution. The conductivity of ionic solutions owes to the presence of ions in solutions. Molecular solutions do not conduct electricity due to the absence of ions.
Glucose is a molecular substance hence they are are no ions present hence glucose does not conduct electricity.
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Suppose you are performing a precipitation titration to study the K sp of zinc(II) iodate, Z n ( I O 3 ) 2 . The literature value of K sp is 3.9 × 10 − 6 . If you use 0.200 M K I O 3 as the titrant, what concentration of I O 3 − (in M) do you predict will be needed to start precipitation in a 0.229 M solution of Z n ( N O 3 ) 2 ?
The predicted concentration of IO₃⁻ needed to start the precipitation titration is 1.703 × 10⁻⁷ M
The dissociation of zinc (II) iodate can be expressed as:
[tex]\mathbf{Zn (IO_3)_2 \to Zn^{2+} + 2IO_3^-}[/tex]
Given that the solubility product constant Ksp value = 3.9 × 10⁻⁶
For the above dissociation,
[tex]\mathbf{Ksp = [Zn^{2+ }] [IO_3^-]^2}[/tex]
Since [tex]\mathbf{ [Zn^{2+ }] = [Zn(NO_3)_2] = 0.229 \ M}[/tex]
∴
[tex]\mathbf{3.9\times 10^{-8} =(0.229) \times [IO_3^-]^2}[/tex]
[tex]\mathbf{ [IO_3^-]^2 = \dfrac{3.9\times 10^{-8} }{(0.229)}}[/tex]
[tex]\mathbf{ [IO_3^-]^2 = 1.703 \times 10^{-7} \ M}[/tex]
Therefore, we can conclude that the initial predicted concentration of IO₃⁻ needed to start the precipitation titration is 1.703 × 10⁻⁷ M
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How many atoms can be found in a sample of 75g of silicon
Answer:
[tex]\boxed {\boxed {\sf 1.6 \times 10^{24} \ atoms \ Si}}[/tex]
Explanation:
We are asked to find how many atoms are in a sample of 75 grams of silicon.
1. Convert Grams to MolesFirst, we must convert grams to moles using the molar mass. This is the mass of 1 mole of a substance. The molar mass is found on the Periodic Table because it is equal to the atomic mass, but the units are grams per mole instead of atomic mass units.
Look up the molar mass of silicon.
Si: 28.085 g/molWe convert using dimensional analysis, so we must set up a conversion factor.
[tex]\frac {28.085 \ g \ Si}{1 \ mol \ Si}[/tex]
We are converting 75 grams of silicon to moles, so we multiply by this value.
[tex]75 \ g \ Si*\frac {28.085 \ g \ Si}{1 \ mol \ Si}[/tex]
Flip the conversion factor so the units of grams of silicon cancel.
[tex]75 \ g \ Si*\frac {1 \ mol \ Si}{28.085 \ g \ Si}[/tex]
[tex]75 *\frac {1 \ mol \ Si}{28.085 }[/tex]
[tex]\frac {75}{28.085} \ mol \ Si[/tex]
[tex]2.670464661 \ mol \ Si[/tex]
2. Convert Moles to AtomsNext, we convert moles to atoms using Avogadro's Number, or 6.022 ×10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this problem, the particles are atoms of silicon.
Set up another conversion factor, this time with Avogadro's Number.
[tex]\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}[/tex]
Multiply by the number of moles we found.
[tex]2.670464661 \ mol \ Si*\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}[/tex]
The units of moles of silicon cancel.
[tex]2.670464661 *\frac {6.022 \times 10^{23} \ atoms \ Si}{1 }[/tex]
[tex]2.670464661 * {6.022 \times 10^{23} \ atoms \ Si}[/tex]
[tex]1.60815382 \times 10^{24} \ atoms \ Si[/tex]
The original value of grams (75) has 2 significant figures, so our answer must have the same. For the number we found, that is the tenths place. The 0 in the hundredths place tells us to leave the 6 in the tenths place.
[tex]1.6 \times 10^{24} \ atoms \ Si[/tex]
The collision theory state that?
Answer:
Collision theory states that molecules must collide to react. For most reactions, however, only a small fraction of collisions produce a reaction.
what is another extraction that uses gravity filtration and describe it
Answer -ˋˏ ༻༺ ˎˊ-
A common use for gravity filtration is for separating anhydrous magnesium sulfate (MgSO4) from an organic solution that it has dried (Figure 1.68b). Anhydrous magnesium sulfate is powdery, and with swirling in an organic solvent creates a fine dispersal of particles like a snow globe.
What is the percent composition of Be(OH)2?
Answer:
Ba 80.148%. H 1.77% O 18.657%
how to experiment titration
Answer:
Add a volume of a solution of known concentration is added to a volume of another solution in order to determine its concentration. Solutions in which a few drops of phenolphthalein have been added turn from colorless to brilliant pink as the solution turns from acidic to basic.
If you want the method:
Method
1) Use a pipette and pipette filler to add 25 cm3 of alkali solution to a clean conical flask.
2) Add a few drops of a suitable indicator and put the conical flask on a white tile.
3) Fill the burette with dilute acid. Flush the tap through to remove any air bubbles. Ensure the burette is vertical.
4) Slowly add the acid from the burette to the conical flask, swirling to mix. (The mixture may at first change colour, and then back again when swirled.)
5) Stop adding the acid when the end-point is reached (when the colour first permanently changes). Note the final volume reading.
6) Repeat steps 1 to 5 until three results are repeatable (in close agreement). Ideally these should lie within 0.10 cm3 of each other.
I hope it helps.
Answer:
To conduct a titration experiment, first fill the burette with an acid or base solution of known concentration. After that, take a burette reading from the top of the miniscus down to the bottom. Then, underneath the burette, place a flask containing an unknown concentration of acid or base. After that, fill the flask halfway with the appropriate indicator and shake it up. Add your titrate to the flask one drop at a time while stirring constantly. Continue to add the titrate until the color change is noticeable. Finally, take one more look at the burette to make sure everything is correct.
Explanation:
Hope it helps:)
ethylene glycol, an antifreeze boils at 197 ⁰C. Convert 197 ⁰C to:
⁰F =
K =
[tex]\boxed{\sf °F=\dfrac{9}{5}°C+32}[/tex]
[tex]\\ \sf\longmapsto °F=\dfrac{9}{5}(197)+32[/tex]
[tex]\\ \sf\longmapsto °F=\dfrac{1773}{5}+32[/tex]
[tex]\\ \sf\longmapsto °F=354.6+32[/tex]
[tex]\\ \sf\longmapsto °F=386.6°F[/tex]
[tex]\rule{200pt}{5pt}[/tex]
[tex]\boxed{\sf K=°C+273}[/tex]
[tex]\\ \sf\longmapsto K=197+273[/tex]
[tex]\\ \sf\longmapsto K=470K[/tex]
Stems are important plant structures because they _________.
A.
are the main sites of photosynthesis for plants
B.
are the only site of water and nutrient transport
C.
allow plants to absorb water and nutrients from the soil
D.
give the plant structure and support
Answer:
c
Explanation:
allow plants to absorb water and nutrients from the soil
Help pretty please :)
How are mass and weight alike?
A. Both change with gravitational force.
B. Both are measured in pounds.
C. Both depend on how much matter an object has.
D. Both are the same everywhere in the universe.
Answer:
C. Both depend on how much matter an object has.
Explanation:
weight is the gravitational force acting on an object, so it does vary depending on gravity.
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What is the reaction between CCl4 and H2O?
Answer:
At temperatures > 400 °C, CCl4 reacts with H2O over a MgO catalyst to yield HCl and CO2.
A carrot originally has a mass of 0.39g and after being left in a salt solution overnight it gained 0.3g. Calculate the percentage mass increase of the carrot.
Answer:
77%
Explanation:
The formula for finding percent change is change/original.
So, you have to do 0.3/0.39
You have to divide this to get 0.769 which can be rounded 0.77
0.77 as a percentage is 77%
So, the percentage mass increase of the carrot is 77%
The partial pressure of CO2 gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO2 gas (in g) will be released from 1.1 L of the carbonated water when the partial pressure of CO2 is lowered to 1.28 atm? At 25 ºC, the Henry’s law constant for CO2 dissolved in water is 1.65 x 103 atm, and the density of water is 1.0 g/cm3.
If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.
The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. We can calculate the concentration of CO₂ using Henry's law.
[tex]C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 4.60 atm = 7.59 \times 10^{-3} M[/tex]
We can calculate the mass of CO₂ in 1.1 L considering its molar mass is 44.01 g/mol.
[tex]\frac{7.59 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.367 g[/tex]
Now, we will repeat the same procedure for a partial pressure of 1.28 atm.
[tex]C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 1.28 atm = 2.11 \times 10^{-3} M[/tex]
[tex]\frac{2.11 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.102 g[/tex]
The mass of CO₂ released will be equal to the difference in the masses at the different pressures.
[tex]m = 0.367 g - 0.102 g = 0.265 g[/tex]
If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.
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The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO₂ gas (in g) will be released from 1.1 L of the carbonated water when the partial pressure of CO2 is lowered to 1.28 atm? At 25 ºC, the Henry’s law constant for CO₂ dissolved in water is 1.65 x 10⁻³ M/atm, and the density of water is 1.0 g/cm³.
A containing vessel holds a gaseous mixture of nitrogen and butane. Thepressure in the vessel at 126.9 Cis 3.0 atm. At 0 C, the butane completelycondenses and the pressure drops to 1.0 atm. Calculate the mole fraction of nitrogenin the original gaseous mixture.
A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.
A vessel contains a gaseous mixture of nitrogen and butane. At 126.9 °C (400.1 K) the pressure is due to the mixture is 3.0 atm.
We can calculate the total number of moles using the ideal gas equation.
[tex]P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{3.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.091 mol/L \times V[/tex]
At 0 °C (273.15 K), the pressure due to the gaseous nitrogen is 1.0 atm.
We can calculate the moles of nitrogen using the ideal gas equation.
[tex]P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{1.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.030 mol/L \times V[/tex]
The mole fraction of nitrogen in the mixture is:
[tex]X(N_2) = \frac{0.030 mol/L \times V}{0.091 mol/L \times V} = 0.33[/tex]
A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.
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Elements that form diatomic molecules include
a oxygen
b hydrogen
C sodium
d two of the above
Answer:
D, two of the above, oxygen and hydrogen.
Explanation:
To solve this problem we can either use orbital stuff or common knowledge. Oxygen and hydrogen are known to form O2 and H2, while sodium forms metallic bonds between the atoms.
To prepare 250.0 mL of 2.5 M KCl you will need to dilute _____ mL of 8.0 M KCl solution to a volume of 250.0 mL.
Answer:
78.125ml
Explanation:
Number of moles in 250ml of 2.5M KCl is (250÷1000)litres×2.5M so we divide these moles by 8M. The answer gotten will be in litres so multiply by 1000 to get it in ml
A 32.1 mole sample of gas has a temperature of 21 Celsius and a pressure of 2280.0 torr. What is the volume of the gas?
The volume of the gas is 27 ml
The calculation can be done as follows
The first step is to convert the pressure to atm
convert 2280 tors to atm
760 tors= 1 atm
= 2280/780
= 3 atm
The formula is
PV= nRT
The next step is to write out the parameters
Pressure= 3 atm
Temperature= 21 °C
no of moles= 32.1 mole
volume= ?
(r)constant= 8.314
3 × v= 32.1×8.314×21
3v= 81.080
V= 81.080/3
V= 27 ml
Hence the volume of the gas is 27 ml
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leopard has canine teeth because
Explanation:
Leopards have 32 teeth, 4 of which are long, pointed canine teeth. Canine teeth are used to kill prey. Other teeth are for cutting flesh and grinding bone
Write the formula of the conjugate base of C6H5SH.
Answer:
C6H5S^-
Explanation:
This is because you are removing the hydrogen to make it a conjugate base.
The formula for conjugate base of C6H5SH is C6H5S⁻.
What is conjugate base?Conjugate base is defined as the substance which is formed when an acid liberate its protons.
The acid donate protons and the acid changes into base.
Conjugate acid is defined as the pair of compound that differs by its protons by gaining protons.
Conjugate acid and base are based in acid base theory.
Thus, the formula for conjugate base of C6H5SH is C6H5S⁻.
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Compare the average motion of the particles in the 3 containers of water
Answer:
c>b=a
Explanation:
It is important to note that mass does not affect the average motion/energy per molecule, but temperature does. the higher the temperature the faster the particles are. A has the same temperature as B, so they have the same amount of motion. C is warmer than A and B, so the average motion of the particles in beaker C is the largest
Air is a mixture of (mostly) oxygen (molecular mass 16) and nitrogen (molecular mass 14) gases. At room temperature, which molecules in this room have more kinetic energy (on average)
The molecules of the two gases will have the same kinetic energy at room temperature.
The average kinetic energy of gaseous molecules can be calculated using the formula:
Kinetic Energy = 3/2RT, where R = constant and T = temperature in Kelvin
This means that the kinetic energy of a gaseous molecule is dependent on the temperature of the molecule only.
In other words, the molecular mass of molecules of gases has no bearing on the kinetic energy of each molecule.
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A city carrier resigns and trains as an office receptionist this is a person’s metabolism is likely to
How many atoms are in mercury phosphate
Answer:
There are 95 atoms in mercury phosphate
Explanation:
You just add the atomic number for both elements
what is chemical reaction ?
Calculate the freezing point of a solution of 3.46 g of a compound, X, in 160 g of benzene. When a separate sample of X was vaporised, its density was found to be 3.27 g/Lat 116°C and 773 torr. The freezing point of pure benzene is 5.45°C, and Kis 5.12°C kg/mol.
Answer:
Calculate the freezing point of a solution of 3.46 g of a compound,X,in 160 g of benzene. When separate sample of X was vaporised, its density was found 3.27 g/L at 116 c and 773 torr. The freezing point of pure benzen is 5.45 c, and kf is 5 .12 c/m.
Explanation:
First determine molar mass from
P*molar mass = density*R*T
Solve for molar mass. Then
moles = grams/molar mass
Solve for moles
molality = moles/kg solvent
Solve for molality
delta T = Kf*m
Solve for delta T, then convert to freezing point