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ss7ja [257]
3 years ago
7

The density of solid Cr is 7.15 g/cm3. How many atoms are present per cubic centimeter of Cr?

Chemistry
1 answer:
Masja [62]3 years ago
8 0
To calculate the number of atoms of Cr, we first find the number of moles per unit of cubic centimeter of Cr. Then, use avogadros number for the number of atoms. Calculations are as follows:

1 cm^3 (7.15 g/cm^3) (1 mol / 51.996 g Cr) = 0.14 mol Cr

0.14 mol Cr ( 6.022 x 10^23 atoms Cr / 1 mol Cr ) = 8.28 x 10^22 atoms Cr
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If the density of water is 1g/cm3 what is its specific gravity
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Specific gravity is the density of asubstance divided by the density of water. Since (at standard temperature and pressure) water has a density of 1 gram/cm3, and since all of the units cancel, specific gravity is usually very close to the same value as density(but without any units).

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What is the molar mass of a gas if a flask with a volume of 3. 16 l contains 9. 33 grams of the gas at 32. 0°c and 1. 00 atm?
Serhud [2]

The molar mass of a gas if a flask with a volume of 3. 16 L contains 9. 33 grams of the gas at 32. 0°C and 1. 00 atm is  1.17g/mol

Calculation ,

In this question we have to fist find the number of moles of gas by using ideal gas equation and from the help of number of moles we can determine molar mass.

According to ideal gas equation which is also known as ideal law ,

PV = nRT                ...( i )

where P is the pressure of the gas = 1 atm

V is the volume of the gas in the flask with volume =  3. 16 L

R is the universal gas constant = 0.082 atm L/K mol

T is the temperature = 32. 0°C = 32 + 273 = 305 K

n is the number of moles = ?

Putting the value of Pressure P , volume V , temperature T , number of moles n and universal gas constant R in the equation (i) we get ,

1 atm ×3. 16 L = n× 0.082 atm L/K mol ×305 K

n = 1 atm ×3. 16 L / 0.082 atm L/K mol × 305 K = 0.126 mole

number of mole of a gas  = 0.126 mole = given mass/ molar mass

molar mass  = number of moles × Given mass =  0.126 × 9. 33 = 1.17g/mol

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1 year ago
What is the result when earth layers experience extension stress? Question 2 options:
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stretching and thinning of the Earth's crust occurs.

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The result of the extension stress on Earth layers is the stretching and thinning of the Earth's crust occurs.

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3 years ago
Determine the molar mass of a 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm. group of answer choices
lilavasa [31]

Considering the ideal gas law and the definition of molar mass, the molar mass of the sample of gas is 9.17 \frac{g}{mol}.

<h3>Ideal gas law</h3>

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

  • P is the gas pressure.
  • V is the volume that occupies.
  • T is its temperature.
  • R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
  • n is the number of moles of the gas.

<h3>Definition of molar mass</h3>

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

<h3>Molar mass of the sample of gas</h3>

In this case you know:

  • P= 0.980 arm
  • V= 1.20 L
  • T= 287 K
  • R= 0.082 \frac{atmL}{molK}
  • n= ?

Replacing in the ideal gas law:

0.980 atm× 1.20 L= n× 0.082\frac{atmL}{molK}× 287 K

Solving:

(0.980 atm× 1.20 L)÷ (0.082\frac{atmL}{molK}× 287 K)= n

<u><em>0.04997 moles= n</em></u>

On the other hand, you know that the<u><em> mass of the sample of gas</em></u> is <u><em>0.458 grams</em></u>. Replacing in the definition of molar mass:

molar mass=\frac{0.458 grams}{0.04997 moles}

Solving:

<u><em>molar mass= 9.17 </em></u>\frac{g}{mol}

Finally, the molar mass of the sample of gas is 9.17 \frac{g}{mol}.

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6 0
2 years ago
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