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3 years ago

6

What is the temperature of 0.620 mol of gas at a pressure of 1.5 atm and a volume of 10.9 L ? Express your answer using two sign

ificant figures.

1 answer:

Sveta_85 [38]3 years ago

6 0

Answer:

320K

Explanation:

PV=nRT

1.5atm(10.9L)=(0.620mol)(0.08206 L atm/ mol K)T

T=320K

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<u>Answer:</u> The mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

<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)

<u>For nickel:</u>

Given mass of nickel = 14.8 g

Molar mass of nickel = 58.7 g/mol

Putting values in equation 1, we get:

$\text{Moles of nickel}=\frac{14.8g}{58.7g/mol}=0.252mol$

For the given chemical reaction:

$3NiO(s)+2Al(s)\rightarrow 3Ni(l)+Al_2O_3(s)$

<u>For nickel (II) oxide:</u>

By Stoichiometry of the reaction:

3 moles of nickel are produced from 3 moles of nickel (II) oxide

So, 0.252 moles of nickel will be produced from $\frac{3}{3}\times 0.252=0.252mol$ of nickel (II) oxide

Now, calculating the mass of nickel (II) oxide by using equation 1:

Molar mass of nickel (II) oxide = 74.7 g/mol

Moles of nickel (II) oxide = 0.252 moles

Putting values in equation 1, we get:

$0.252mol=\frac{\text{Mass of nickel (II) oxide}}{74.7g/mol}\\\\\text{Mass of nickel (II) oxide}=(0.252mol\times 74.7g/mol)=18.8g$

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By Stoichiometry of the reaction:

3 moles of nickel are produced from 2 moles of aluminium

So, 0.252 moles of nickel will be produced from $\frac{2}{3}\times 0.252=0.168mol$ of aluminium

Now, calculating the mass of aluminium by using equation 1:

Molar mass of aluminium = 27 g/mol

Moles of aluminium = 0.168 moles

Putting values in equation 1, we get:

$0.168mol=\frac{\text{Mass of aluminium}}{27g/mol}\\\\\text{Mass of aluminium}=(0.168mol\times 27g/mol)=4.54g$

Hence, the mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

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nirvana33 [79]

Answer:

The answer to your question is P2 = 2676.6 kPa

Explanation:

Data

Volume 1 = V1 = 12.8 L Volume 2 = V2 = 855 ml

Temperature 1 = T1 = -108°C Temperature 2 = 22°C

Pressure 1 = P1 = 100 kPa Pressure 2 = P2 = ?

Process

- To solve this problem use the Combined gas law.

P1V1/T1 = P2V2/T2

-Solve for P2

P2 = P1V1T2 / T1V2

- Convert temperature to °K

T1 = -108 + 273 = 165°K

T2 = 22 + 273 = 295°K

- Convert volume 2 to liters

1000 ml -------------------- 1 l

855 ml -------------------- x

x = (855 x 1) / 1000

x = 0.855 l

-Substitution

P2 = (12.8 x 100 x 295) / (165 x 0.855)

-Simplification

P2 = 377600 / 141.075

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P2 = 2676.6 kPa

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3 years ago