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Bogdan [553]
3 years ago
11

Calculate the mass of 120 cc of nitrogen at STP.How many

Chemistry
2 answers:
german3 years ago
6 0

Answer:

6.02 x 10²⁰ molecules

Explanation:

Given parameters:

Volume of nitrogen  = 120cm³

Unknown:

Mass of Nitrogen = ?

Number of molecules present in it = ?

Solution:

At STP;

       1 mole of a substance contains 22.4dm³ particles

      100cm³ = 1dm³

     24cm³ will give 0.024dm³

        22.4dm³ of particles are in 1 mole of substances at STP

         0.024dm³ of particles will contain  \frac{0.024}{24}   = 0.001mole

Mass of Nitrogen  = Number of moles x molar mass

  molar mass of nitrogen  = 14g/mol

   Mass of nitrogen  = 0.001 x 14  = 0.014g

Number of molecules present;

      1 mole of a substance contains 6.02 x 10²³ molecules

     0.001 mole of Nitrogen will contain 0.001  x  6.02 x 10²³ = 6.02 x 10²⁰

         6.02 x 10²⁰ molecules

Korvikt [17]3 years ago
3 0

2.67 × 10^19

Explaination:

The volume of the gas is 1c.c = 10^ - 3L

∴ Number of moles of the gas is=22.4 × 10^-3

Now,

Number of molecules = n × 6.022 × 10^23

∴ Number of Nitrogen Molecules

= 10^-3/22.4 × 6.023 × 10^23

  = 2.67 × 10^19

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12.5 g of copper are reacted with an excess of chlorine gas, and 25.4 g of copper(II) chloride are
Alika [10]

Answer:

Percent yield = 94.5%

Theoretical yield =  26.89 g

Explanation:

Given data:

Mass of copper = 12.5 g

Mass of copper chloride produced = 25.4 g

Theoretical yield = ?

Percent yield = ?

Solution:

Cu + Cl₂  →  CuCl₂

Number of moles of Copper:

Number of moles = mass/ molar mass

Number of moles = 12.5 g/ 63.55 g/mol

Number of moles = 0.2 mol

Now we will compare the moles of copper with copper chloride.

          Cu          :           CuCl₂

           1             :              1

          0.2          :            0.2

Theoretical yield:

Mass of copper chloride:

Mass = Number of moles × molar mass

Mass = 0.2 mol × 134.45 g/mol

Mass = 26.89 g

Percent yield:

Percent yield = Actual yield / theoretical yield  × 100

Percent yield = 25.4 g/26.89 g × 100

Percent yield = 94.5%

8 0
4 years ago
What is the mass of 19.5 moles of carbon dioxide?
Olenka [21]

Explanation:

mass = moles * molar mass

molar mass of carbon dioxide (CO2) = 12+32 = 44

= 19.5* 44 = 858

proof

moles = mass /molar mass

= 858/44 = 19.5 moles

6 0
3 years ago
What happens to the atoms and molecules in a substance as its temperature increases?
lana [24]
The molecules and atoms vibrate faster. As atoms vibrate faster the space between atoms also increases.
5 0
3 years ago
A compound is composed of only C, H, and O. The combustion of a .519-g sample of the compound yields 1.24 g of CO_2 and .255 g o
Kay [80]

Answer:

Empirical formula is C₃H₃O.

Explanation:

Given data:

Mass of compound = 0.519 g

Mass of CO₂ = 1.24 g

Mass of H₂O = 0.255 g

Empirical formula = ?

Solution:

%age of C,H,O

C = 1.24 g/0.519 × 12/44 ×100 = 65.5%

H = 0.255 g/0.519 × 2.016/18 ×100 =  5.6%

O = 100 - (65.5+5.6)

O = 28.9%

Number of gram atoms of H = 5.6 / 1.01 = 5.5

Number of gram atoms of O =  28.9 / 16 = 1.81

Number of gram atoms of C = 65.5 / 12 = 5.5

Atomic ratio:

            C                      :        H              :         O

          5.5/1.81              :       5.5/1.81     :       1.81/1.81

            3                      :        3              :        1

C : H : O = 3 : 3 : 1

Empirical formula is C₃H₃O.

8 0
4 years ago
ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J.
timofeeve [1]

Complete question:

ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J. calculate the magnitude of q for the process.

Answer:

The magnitude of q for the process 568 J.

Explanation:

Given;

change in internal energy of the gas, ΔU = 475 J

work done by the gas, w = 93 J

heat added to the system, = q

During gas expansion process, heat is added to the gas.

Apply the first law of thermodynamic to determine the magnitude of heat added to the gas.

ΔU = q - w

q = ΔU +  w

q = 475 J  +  93 J

q = 568 J

Therefore, the magnitude of q for the process 568 J.

6 0
3 years ago
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