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2 years ago

What is the volume of 1.56 kg of a compound whose molar mass is 81.86 g/mole and whose density is 41.2 g/ml?

Chemistry

1 answer:

hjlf2 years ago

7 0

Answer:

v = 37.9 ml

Explanation:

Given data:

Mass of compound = 1.56 kg

Density = 41.2 g/ml

Volume of compound = ?

Solution:

First of all we will convert the mass into g.

1.56 ×1000 = 1560 g

Formula:

D=m/v

D= density

m=mass

V=volume

v = m/d

v = 1560 g / 41.2 g/ml

v = 37.9 ml

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PLEASE ANSWER I AM BEGGING

TiliK225 [7]

Taking into account the definition of dilution:

you have to use 8.23 mL of a stock solution of 7.00 M HNO₃ to prepare 0.120 L of 0.480 M HNO₃.
If you dilute 20.0 mL of the stock solution to a final volume of 0.270 L , the concentration of the diluted solution is 0.518 M.

<h3>Dilution</h3>

Dilution is the process of reducing the concentration of solute in solution, which is accomplished by simply adding more solvent to the solution at the same amount of solute.

In a dilution the amount of solute does not change, but as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.

A dilution is mathematically expressed as:

Ci×Vi = Cf×Vf

where

Ci: initial concentration
Vi: initial volume
Cf: final concentration
Vf: final volume

<h3>Volume of stock solution</h3>

In this case, you know:

Ci= 7 M
Vi= ?
Cf= 0.480 M
Vf= 0.120 L

Replacing in the definition of dilution:

7 M× Vi= 0.480 M× 0.120 L

Solving:

Vi= (0.480 M× 0.120 L)÷ 7 M

<u><em>Vi= 0.00823 L= 8.23 mL</em></u> (being 1 L= 1000 mL)

Finally, you will need 8.23 mL of the stock solution.

<h3>Concentration of the diluted solution</h3>

In this case, you know:

Ci= 7 M assuming the stock solution is 7.00 M HNO₃
Vi= 20 mL= 0.02 L
Cf= ?
Vf= 0.270 L

Replacing in the definition of dilution:

7 M× 0.02 L= Cf× 0.270 L

Solving:

(7 M× 0.02 L)÷ 0.270 L= Cf

Finally, the concentration is 0.518 M.

Learn more about dilution:

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5 0

1 year ago

A pycnometer is a precisely weighted vessel that is used for highly accurate density determinations. Suppose that a pycnometer h

dexar [7]

Answer:

5.758 is the density of the metal ingot in grams per cubic centimeter.

Explanation:

1) Mass of pycnometer = M = 27.60 g

Mass of pycnometer with water ,m= 45.65 g

Density of water at 20 °C = d = $998.2 kg/m^3$

1 kg = 1000 g

$1 m^3=10^6 cm^3$

$998.2 kg/m^3=\frac{998.2 \times 1000 g}{10^6 cm^3}=0.9982 g/cm^3$

Mass of water ,m'= m - M = 45.65 g - 27.60 g =18.05 g

Volume of pycnometer = Volume of water present in it = V

$Density=\frac{Mass}{Volume}$

$V=\frac{m'}{d}=\frac{18.05 g}{0.9982 g/cm^3}=18.08 cm^3$

2) Mass of metal , water and pycnometer = 56.83 g

Mass of metal,M' = 9.5 g

Mass of water when metal and water are together ,m''= 56.83 g - M'- M

56.83 g - 9.5 g - 27.60 g = 19.7 g

Volume of water when metal and water are together = v

$v=\frac{m''}{d}=\frac{19.7 g}{0.9982 g/cm^3}=19.73 cm^3$

Density of metal = d'

Volume of metal = v' = $\frac{M'}{d'}$

Difference in volume will give volume of metal ingot.

v' = v - V

$v'=19.73 cm^3-18.08 cm^3=$

$v'=1.65 cm^3$

Since volume cannot be in negative .

Density of the metal =d'

= $d'=\frac{M'}{v'}=\frac{9.5 g}{1.65 cm^3}=5.758 g/cm^3$

5 0

2 years ago

Write the equation for the formation of water gas

GREYUIT [131]

Answer:

the answer is C + H2O # CO + H2.

4 0

2 years ago

Read 2 more answers

Which property of matter changes during a chemical change but not during a physical change

andrew-mc [135]

Answer:In a physical change, atoms are not rearranged and the matter's physical and chemical properties are unchanged. Chemical changes, on the other hand, rearrange the atoms of matter in new combinations, resulting in matter with new physical and chemical properties.

Explanation:

easy

8 0

3 years ago

Explain, in detail, how you convert grams of one substance to grams of something else. Be specific and include each step

elena-14-01-66 [18.8K]

I will present a simple reaction so we can do this conversion:

2H₂ + O₂ → 2H₂O

We will assume we have 32 g of O₂ and we want to find the amount of water, assuming this reaction goes to completion. We must first convert the initial mass to moles, which we do using the molar mass in units of g/mol. The molar mass of O₂ is 32 g/mol.

32 g O₂ ÷ 32 g/mol = 1 mole O₂.

Now that we have moles of oxygen, we use the molar coefficients to find the ratio of water molecules to oxygen molecules. We can see there are 2 moles of water for every 1 mole of oxygen.

1 moles O₂ x (2 mol H₂O/ 1 mol O₂) = 2 moles H₂O

Now that we have the moles of water, we can convert this amount into grams using the molar mass of water, which is 18 g/mol.

2 moles H₂O x 18 g/mol = 36 g H₂O

Now we have successfully converted the mass of one molecule to the mass of another.

5 0

3 years ago