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2 years ago

What is the approximate partial pressure of oxygen at 1500 m?

Chemistry

1 answer:

elena55 [62]2 years ago

3 0

The approximate partial pressure of oxygen at 1500 m is 0.18 atm.

<h3>Pressure of the air</h3>

From barometric pressure table;

1500 m = 85 kPa

1 kPa = 0.0099 atm

85 kPa = ?

= 0.84 atm

Oxygen percentage in air = 21%

<h3>Partial pressure of oxygen in air</h3>

P = 0.21 x P(total)

P = 0.21 x 0.84 atm

P = 0.18 atm

Thus, the approximate partial pressure of oxygen at 1500 m is 0.18 atm.

Learn more about partial pressure here: brainly.com/question/19813237

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KIM [24]

Answer: 24 moles of $SnF_2$ are produced.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

${\text{Moles of} H_2}=\frac{48g}{2g/mol}=24moles$

$Sn+2HF\rightarrow SnF_2+H_2$

According to stoichiometry :

1 mole of $H_2$ is accompanied with = 1 mole of $SnF_2$

Thus 24 moles of $H_2$ is accompanied with = $\frac{1}{1}\times 24=24moles$ of $SnF_2$

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25 g of a compound is added to 500 mL of water if the freezing point of the resulting solution is

vlabodo [156]

Answer:

a) 119 g/mol

Explanation:

-We apply the formula for freezing point depression to obtain the molality of the solution:

$\bigtriangleup T_f=K_fm, \ \ K_f=1.36\textdegree C/m\\\\\therefore m=\frac{\bigtriangleup T_f}{K_f}\\\\=\frac{0.57\textdegree C}{1.36\textdegree C}\\\\=0.4191\ mol/Kg\\\\$

#We use the molality above to calculate the molar mass:

$m=\frac{0.4191\ mol}{1\ Kg}=\frac{25\ g}{0.5\ Kg}\\\\\therefore 1 \ mol=\frac{25\ g}{0.5\ Kg}\times\frac{1\ Kg}{ 0.4191}\\\\=119.3033\approx 119\ g/mol$