3 years ago

Will mark brainliest! Topic is redox reactions, please follow the insturctions given above to help

Chemistry

1 answer:

TiliK225 [7]3 years ago

3 0

Answer:

do you really need help or are you not paying attention in class

Explanation:

also i cant dee it well

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You have 8 moles of a gas at 250 K in a 6 L container. What is the pressure of the gas? Show your work

Bogdan [553]

Hello:

In this case, we will use the Clapeyron equation:

P = ?

n = 8 moles

T = 250 K

R = 0.082 atm.L/mol.K

V = 6 L

Therefore:

P * V = n * R * T

P * 6 = 8 * 0.082* 250

P* 6 = 164

P = 164 / 6

P = 27.33 atm

Hope that helps!

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Mg+O2=MgO
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Under certain conditions Argon gas diffuses at a rate of 3.2 cm per second under the same conditions an unknown gas diffuses at

kozerog [31]

Answer:

20 g/mol

Explanation:

We can use Graham’s Law of diffusion:

The rate of diffusion (r) of a gas is inversely proportional to the square root of its molar mass (M).

$r = \frac{1 }{\sqrt{M}}$

If you have two gases, the ratio of their rates of diffusion is

$\frac{r_{2}}{r_{1}} = \sqrt{\frac{M_{1}}{M_{2}}}$

Squaring both sides, we get

$(\frac{r_{2}}{r_{1}})^{2} = \frac{M_{1}}{M_{2}}$

Solve for M₂:

$M_{2} = M_{1} \times (\frac{r_{1}}{r_{2}})^{2}$

$M_{2} = \text{39.95 g/mol} \times (\frac{\text{3.2 cm/s}}{\text{4.5 cm/s}})^{2}= \text{39.95 g/mol} \times (0.711 )^{2}$

$= \text{39.95 g/mol} \times 0.506 = \textbf{20 g/mol}$

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3 years ago