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3 years ago

H2O (s) ⇒ H2O What would be the best argument against writing this in the form of a chemical equation?

Chemistry

1 answer:

Goshia [24]3 years ago

5 0

Answer: Since this is merely a Physical Change, it is improper to write it as a chemical equation.

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Consider a mixture of two gases, A and B, confined in a closed vessel. A quantity of a third gas, C, is added to the same vessel

n200080 [17]

The question is incomplete, here is the complete question:

Consider a mixture of two gases, A and B, confined in a closed vessel. A quantity of a third gas, C, is added to the same vessel at the same temperature. How does the addition of gas C affect the following. The mole fraction of gas B?

A mixture of gases contains 10.25 g of N₂, 2.05 g of H₂, and 7.63 g of NH₃.

Answer: The mole fraction of gas B (hydrogen gas) is 0.557

Explanation:

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

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

For nitrogen gas:

Given mass of nitrogen gas = 10.25 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen gas}=\frac{10.25g}{28g/mol}=0.366mol$

For hydrogen gas:

Given mass of hydrogen gas = 2.05 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrogen gas}=\frac{2.05g}{2g/mol}=1.025mol$

For ammonia gas:

Given mass of ammonia gas = 7.63 g

Molar mass of ammonia gas = 17 g/mol

Putting values in equation 1, we get:

$\text{Moles of ammonia gas}=\frac{7.63g}{17g/mol}=0.449mol$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B+n_C}$

Moles of gas B (hydrogen gas) = 1.025 moles

Total moles = [0.366 + 1.025 + 0.449] = 1.84 moles

Putting values in above equation, we get:

$\chi_{(H_2)}=\frac{1.025}{1.84}=0.557$

Hence, the mole fraction of gas B (hydrogen gas) is 0.557

6 0

3 years ago

A buffer is made that is .10M propanoic acid (pKa = 4.9) and .05M sodium propanoate. Calculate the pH.

inna [77]

PKa= 4.9 therefore ka= 10^-4.9= 1.259x10^-5

$ka= \frac{[H^+][CH3CH2COO^-]}{[CH3CH2COOH]}$

$[CH3CH2COO^-]
$ = 0.05
$[CH3CH2COOH]$ = 0.10

Therefore 1.259x10^-5 = $\frac{[H^+][0.05]}{[0.1]}$
Rearrange the equation to make the concentration of hydrogen the subject.
Therefore $[H^+] = \frac{(1.259*10^-5)(0.1)}{0.05}$

Therefore $[H^+]= 2.513*10^-5$

pH= -log [ $H^+$ ] = -log(2.513*10^-5)= 4.59.

7 0

3 years ago

How does weathering break down or change rocks

tatyana61 [14]

Well weathering adds different elements to rocks which they might not be used to. For example, very hard rain & harsh winds could push a rock, causing it to fall & break off into smaller rocks.

3 0

3 years ago

How do i know if a reaction is exo/endothermic?? help please.

goblinko [34]

Answer:

Explanation: in exothermic reaction heat is released out

In endothermic reaction heat is absorbed

3 0

3 years ago

Which of the following is not a required characteristic for a living organism?

m_a_m_a [10]

Answer:

Explanation:

living things are not required to move

8 0

2 years ago

Read 2 more answers