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Liono4ka [1.6K]
2 years ago
14

I) If 8.87g of As2O3 is used in the reaction and 5.33 g of As is produced, what is the percent

Chemistry
1 answer:
Diano4ka-milaya [45]2 years ago
7 0

Answer:

79.3%

Explanation:

Percent yield = 5.33g/6.72g x 100% = 79.3%

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Will a horse galloping in a field produce and average velocity of zero
motikmotik

Answer:

No

Explanation:

Depending on the mass of the horse and the speed, velocity will change.

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An increase in the surface area of reactants in a heterogeneous reaction will result in what?
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It will result in an increase in the rate of rxn
4 0
2 years ago
Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79
BARSIC [14]

Answer :  The mole fraction and partial pressure of CH_4,CO_2 and He gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of CH_4 = 1.79 mole

Moles of CO_2 = 1.20 mole

Moles of He = 3.71 mole

Now we have to calculate the mole fraction of CH_4,CO_2 and He gases.

\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267

and,

\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179

and,

\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554

Thus, the mole fraction of CH_4,CO_2 and He gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of CH_4,CO_2 and He gases.

According to the Raoult's law,

p_i=X_i\times p_T

where,

p_i = partial pressure of gas

p_T = total pressure of gas  = 5.78 atm

X_i = mole fraction of gas

p_{CH_4}=X_{CH_4}\times p_T

p_{CH_4}=0.267\times 5.78atm=1.54atm

and,

p_{CO_2}=X_{CO_2}\times p_T

p_{CO_2}=0.179\times 5.78atm=1.03atm

and,

p_{He}=X_{He}\times p_T

p_{He}=0.554\times 5.78atm=3.20atm

Thus, the partial pressure of CH_4,CO_2 and He gases are, 1.54, 1.03 and 3.20 atm respectively.

4 0
3 years ago
A cell with osmolarity measuring 300 mosm/l is placed in a beaker that contains a nacl solution with osmolarity of 150mosm/l. pr
tia_tia [17]
<span>My hypothesis is the the cell, having a higher osmolarity than the solution of of nacl in the beaker, will have an osmosis reaction releasing into the solution of nacl. This will continue until both cell and solution reach a balance.</span>
5 0
3 years ago
A solution made by mixing 20.0 g of a non-volatile compound with 125 mL of water at 25°C has a vapor pressure of 22.67 torr. Wha
Ainat [17]

We have that the molecular weight (3sf) of the compound (g/mol)

m=44.15g/mol

From the question we are told

A solution made by mixing 20.0 g of a non-volatile compound with 125 mL of water at 25°C has a vapor pressure of 22.67 torr. What is the molecular weight (3sf) of the compound (g/mol).

Generally the equation for the Rouault's law is mathematically given as

P=P_0 N

22.67=23.8*\frac{\frac{12.5}{18}}{\frac{125}{18}+\frac{15}{m}}\\\\\6.95+\frac{15}{m}=7.29\\\\\frac{15}{m}=7.29-6.95\\\\m=\frac{15}{0.34}\\\\m=44.11g/mol

Therefore

The molecular weight (3sf) of the compound (g/mol)

m=44.15g/mol

For more information on this visit

brainly.com/question/17756498

4 0
2 years ago
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