The partial pressure of 0.50 Ne gas is 214.71 torr

calculation

the partial pressure of Ne = moles of Ne/total moles x final pressure

find the total moles of the air mixture

that is moles of Ne + moles of K= 0.50 + 1.20 = 1.70 moles

The partial pressure is therefore = 0.50 /1.70 x 730 = 214.71 torr

8 0

3 years ago

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The elementary reaction 2H2O(g)↽−−⇀2H2(g)+O2(g) 2H2O(g)↽−−⇀2H2(g)+O2(g) proceeds at a certain temperature until the partial pres

Dima020 [189]

Answer:

$6.25\times 10^{-6}$ is the value of the equilibrium constant at this temperature.

Explanation:

Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of products to the partial pressures of reactants each raised to the power equal to their stoichiometric ratios. It is expressed as $K_{p}$

$2H_2O(g)\rightleftharpoons 2H_2(g)+O_2(g)$

Partial pressures at equilibrium:

$p^o_{H_2O}=0.070 atm$

$p^o_{H_2}=0.0035 atm$

$p^o_{O_2}=0.0025 atm$

The equilibrium constant in terms of pressures is given as:

$K_p=\frac{(p^o_{H_2})^2\times (p^o_{O_2})}{(p^o_{H_2O})62}$

$K_p=\frac{(0.0035 atm)^2\times 0.0025 atm}{(0.070 atm)^2}=6.25\times 10^{-6}$

$6.25\times 10^{-6}$ is the value of the equilibrium constant at this temperature.

5 0

3 years ago

What about 50 g of water?<br> I need help what this

ELEN [110]

Answer:

3.38 Tablespoons

10.14 Teaspoons

0.21 U.S. Cups

0.18 Imperial Cups

0.20 Metric Cups

50.00 Milliliters

Explanation:

3 0

3 years ago

How to classify hemicetal and acetal

kumpel [21]

You might have meant hemiacetal, not hemicetal.
Acetals contain two –OR groups, one –R group and a –H atom. In hemiacetals, one of the –OR groups in acetals is replaced by a –OH group<span>.
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3 0

3 years ago