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3 years ago

Which of the following gas samples would have the largest volume, if all samples are at the same temperature and pressure?

Chemistry

1 answer:

crimeas [40]3 years ago

6 0

Answer:

a) 8×10²³ molecules of N2

Explanation:

Let us consider the same temperature and pressure of both sample are standard temperature and pressure.

Standard temperature = 273.15 K

Standard pressure = 1 atm

Number of moles of CO₂ = 0.107 mol

Volume of CO₂ = ?

Solution:

PV = nRT

R = general gas constant = 0.0821 atm.L /mol.K

1 atm × V = 0.107 mol × 0.0821 atm.L /mol.K × 273.15 K

1 atm × V = 2.4 atm.L

V = 2.4 atm.L/1 atm

V = 2.4 L

Volume of N₂:

Number of moles of N₂:

1 mole = 6.022 ×10²³ molecules

8×10²³ molecules × 1 mol / 6.022 ×10²³ molecules

1.33 mol

PV = nRT

R = general gas constant = 0.0821 atm.L /mol.K

1 atm × V = 1.33 mol × 0.0821 atm.L /mol.K × 273.15 K

1 atm × V = 29.83 atm.L

V = 29.83 atm.L/1 atm

V = 29.83 L

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4 years ago

Identify the limiting reactant when 9.65-g H2SO4 reacts with 6.10-g of NaOH. I need help

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3 years ago

A mixture of A and B is capable of being ignited only if the mole percent of A is 6 %. A mixture containing 9.0 mole% A in B flo

atroni [7]

Explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

Weight = $0.09 \times 16.04 + 0.91 \times 29$

= 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

$\frac{800}{27.8336}$

= 28.74 kmol/hr

Now, applying component balance as follows.

$0.09 \times 28.74 + 0 \times F_{B} = 0.06F_{p}$

$F_{p}$ = 43.11 kmol/hr

$F_{A} + F_{B} = F_{p}$

$F_{B}$ = 43.11 - 28.74

= 14.37 kmol/hr

So, mass flow rate of pure (B), is $F_{B}$ = $14.37 \times 29$

= 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

Molecular weight of product stream = $Mol. weight \times 43.11 kmol/hr$

= $0.06 \times 16.04 + 0.94 \times 29$

= 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of $O_{2}$ into the product stream is as follows.

$0.21 \times 0.94 \times 43.11$

= $8.5099 kmol/hr \times 329 g/mol$

= 272.31 kg/hr

Therefore, calculate the mass % of $O_{2}$ into the stream as follows.

$\frac{272.31}{1216.67} \times 100$

= 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of $O_{2}$ in the product gas is 22.38%.

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3 years ago

Which of the following explains why the lab safety policy specifically addresses loose clothing, contact lenses, and large, loos

valentina_108 [34]

Answer:

b. Loose clothing and jewelry is more likely to impede movement, or accidentally knock materials over, causing a spill or other accident.

Explanation:

Hello,

During the laboratories, a high dexterity is required, thus, such accessories, when you move or change your duty, could touch other materials or act out of your control resulting in spillages and constricting your proper movements and skills, that is why considering the lab safety policy, we must take those artifacts away from the chemistry lab.

Best regards.

3 0

3 years ago