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mestny [16]
3 years ago
15

The outside temperature is 35°C. What is the temperature in K

Chemistry
1 answer:
Fittoniya [83]3 years ago
5 0

Answer:

308.15

Explanation:

308.15K − 273.15 = 35°C

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A student has a 2.19 L bottle that contains a mixture of O 2 , N 2 , and CO 2 with a total pressure of 5.57 bar at 298 K . She k
Sergeeva-Olga [200]

<u>Answer:</u> The partial pressure of oxygen gas is 2.76 bar

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 5.57 bar

V = Volume of the gas = 2.19 L

T = Temperature of the gas = 298 K

R = Gas constant = 0.0831\text{ L bar }mol^{-1}K^{-1}

n = Total number of moles = ?

Putting values in above equation, we get:

5.57bar\times 2.19L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{5.57\times 2.19}{0.0831\times 298}=0.493mol

To calculate the mole fraction of carbon dioxide, we use the equation given by Raoult's law, which is:

p_{A}=p_T\times \chi_{A}         ........(1)

where,

p_A = partial pressure of carbon dioxide = 0.318 bar

p_T = total pressure = 5.57 bar

\chi_A = mole fraction of carbon dioxide = ?

Putting values in above equation, we get:

0.318bar=5.57bar\times \chi_{CO_2}\\\\\chi_{CO_2}=\frac{0.381}{5.57}=0.0571

  • Mole fraction of a substance is given by:

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

We are given:

Moles of nitrogen gas = 0.221 moles

Mole fraction of nitrogen gas, \chi_{N_2}=\frac{0.221}{0.493}=0.448

Calculating the partial pressure of oxygen gas by using equation 1, we get:

Mole fraction of oxygen gas = (1 - 0.0571 - 0.448) = 0.4949

Total pressure of the system = 5.57 bar

Putting values in equation 1, we get:

p_{O_2}=5.57bar\times 0.4949\\\\p_{O_2}=2.76bar

Hence, the partial pressure of oxygen gas is 2.76 bar

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Which statement is correct with respect to gross primary productivity
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Codeine C18H21NO3 is a weak organic base. A 5.0 x 10^-3 M solution of codeine has a pH of 9.95. Calculate the value of Kb for th
arsen [322]

Answer:

Kb = 4.45\times10^{-7}\ mol/L

p^{Kb}=6.35

Explanation:

For a weak organic base, the formula to find p^{OH} is given by:

p^{OH}=p^{K_b}+\log c

where c is the concentration of base.

Here c= 5\times10^{-3}\ M

p^{H}=9.95\\p^{OH}=14-p^{H}=14-9.95=4.05

Substituting the above values in the formula,we get:

p^{k_b}=p^{OH}-\log c\\p^{k_b}=4.05-\log (5\times10^{-3})\\p^{K_b}=6.35\\K_b=$antilog 6.35=4.45\times10^{-7}\ mol/L

Hence:

Kb = 4.45\times10^{-7}\ mol/L

p^{Kb}=6.35

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3 years ago
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