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Anuta_ua [19.1K]
3 years ago
8

2. At STP, a sample of gas occupies 24.5 mL. Calculate the volume of this gas at a

Chemistry
1 answer:
Law Incorporation [45]3 years ago
6 0

Answer:

11.7 ml = V

Explanation: Here

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4 years ago
A flexible container at an initial volume of 6.13 L contains 2.51 mol of gas. More gas is then added to the container until it r
aleksley [76]

Answer:

2.12 moles of gas were added.

Explanation:

We can solve this problem by using<em> Avogadro's law</em>, which states that at constant temperature and pressure:

  • V₁n₂=V₂n₁

Where in this case:

  • V₁ = 6.13 L
  • n₂ = ?
  • V₂ = 11.3 L
  • n₁ = 2.51 mol

We <u>input the data</u>:

  • 6.13 L * n₂ = 11.3 L * 2.51 mol
  • n₂ = 4.63

As <em>4.63 moles is the final number of moles</em>, the number of moles added is:

  • 4.63 - 2.51 = 2.12 moles
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3 years ago
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3 years ago
Gasoline burning in an engine: Exothermic or Endothermic?
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8 0
4 years ago
A sample of gallium Bromide GaBr2,weighing 0.165 g was dissolved in water and treated with silver nitrate AgNO3, and resulting t
tresset_1 [31]

<u>Answer:</u> The percent gallium in gallium bromide is 30.30 %.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of gallium bromide = 0.165 g

Molar mass of titanium gallium bromide = 229.53 g/mol

Putting values in equation 1, we get:

\text{Moles of gallium bromide}=\frac{0.165g}{229.53g/mol}=0.00072mol

  • The chemical equation for the reaction of gallium bromide and silver nitrate follows:

GaBr_2+2AgNO_3\rightarrow 2AgBr(s)+Ga(NO_3)_2

By Stoichiometry of the reaction:

1 moles of gallium bromide produces 1 mole of gallium nitrate

So, 0.00072 moles of gallium bromide will produce = \frac{1}{1}\times 0.00072=0.00072moles of gallium nitrate

  • Now, calculating the mass of gallium nitrate from equation 1, we get:

Molar mass of gallium nitrate = 193.73 g/mol

Moles of gallium nitrate = 0.00072 moles

Putting values in equation 1, we get:

0.00072mol=\frac{\text{Mass of gallium nitrate}}{193.73g/mol}\\\\\text{Mass of gallium nitrate}=0.139g

Calculating the mass of gallium in the reaction, we use unitary method:

In 1 mole of gallium nitrate, 1 mole of gallium atom is present.

In 193.73 grams of gallium nitrate, 69.72 g of gallium atom is present.

So, in 0.139 grams of gallium nitrate, the mass of gallium present will be = \frac{69.72}{193.73}\times 0.139=g

  • To calculate the percentage composition of gallium in gallium bromide, we use the equation:

\%\text{ composition of gallium}=\frac{\text{Mass of gallium}}{\text{Mass of gallium bromide}}\times 100

Mass of gallium bromide = 0.165 g

Mass of gallium = 0.050 g

Putting values in above equation, we get:

\%\text{ composition of gallium}=\frac{0.050g}{0.165g}\times 100=30.30\%

Hence, the percent gallium in gallium bromide is 30.30 %.

3 0
3 years ago
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