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4 years ago

what is the pressure of a mixture of co2, so2, and h2o gasses, if each gas has a partial pressure of 25 kpa?

Chemistry

2 answers:

vodka [1.7K]4 years ago

8 0

Answer: 75 kPa

Explanation: According to Dalton's law, the total pressure of a mixture of gases is the sum of individual pressures exerted by the constituent gases.

$p_{total}=p_A+p_B+p_C$

Thus $p_{total}=p_{CO_2}+p_{SO2}+p_{H_2O}$

Given:

$p_{CO_2}=25kPa$

$p_{SO_2}=25kPa$

$p_{H_2O}= 25kPa$

Thus $p_{total}=25Pa+25kPa+25kPa$

$p_{total}=75kPa$

Thus the pressure of mixture of gases is 75 kPa.

VikaD [51]4 years ago

7 0

Are three mixtures:

CO₂ , SO₂ , H₂O so :

25 Kpa x 3 = 75 Kpa

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During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the exce

swat32

Answer: 16.32 g of $O_2$ as excess reagent are left.

Explanation:

To calculate the moles :

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

$\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol$

$\text{Moles of} O_2=\frac{21.71g}{32g/mol}=0.68mol$

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

According to stoichiometry :

2 moles of $SO_2$ require = 1 mole of $O_2$

Thus 0.34 moles of $SO_2$ will require= $\frac{1}{2}\times 0.34=0.17moles$ of $O_2$

Thus $SO_2$ is the limiting reagent as it limits the formation of product and $O_2$ is the excess reagent.

Moles of $O_2$ left = (0.68-0.17) mol = 0.51 mol

Mass of $O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g$

Thus 16.32 g of $O_2$ as excess reagent are left.

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In an electrolytic cell, the magnitude of the standard cell potential is the:______

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What is Polar and Non-polar Covalent bond ?

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3 years ago

Read 2 more answers

Calculate the percentage of each element in acetic acid, hc2h3o2, and glucose, c6h12o6.

My name is Ann [436]

Acetic acid, HC2H3O2

First, calculate for the molar mass of acetic acid as shown below.
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Then, calculating for the percentages of each element.
 Hydrogen:
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 Carbon:
P2 = ((2)(12)/60)(100%) = 40%

Oxygen
P3 =((2)(16) / 60)(100%) = 53.33%

Glucose, C6H12O6

The molar mass of glucose is as calculated below,
6(12) + 12(1) + 6(16) = 180

The percentages of the elements are as follow,
 Hydrogen:
P1 = (12/180)(100%) = 6.67%

Carbon:
P2 = ((6)(12) / 180)(100%) = 40%

Oxygen:
P3 = ((6)(16) / 180)(100%) = 53.33%

b. Since the empirical formula of the given substances are just the same and can be written as CH2O then, the percentages of each element composing them will just be equal.

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