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3 years ago

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A student performs a reaction that makes aluminum oxide. According to her calculations, she should expect to make 115.2 grams. S

he actually produces 66.9 grams. What is her percent yield?

1 answer:

tiny-mole [99]3 years ago

7 0

Answer:

The percentage of the student is 58.17%.

Explanation:

Expected yield of aluminum oxide = 115.2 g

Actual yield of aluminum oxide produced =66.9 g

The percentage yield is calculated by dividing actual yield by expected yield and then multiplying it with hundred.

Percentage yield:

$\frac{\text{Actual yield}}{\text{Expected yield}}\times 100$

$\% Yield=\frac{66.9 g}{115.2 g}\times 100=58.17\%$

The percentage of the student is 58.17%.

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What is the pressure of a sample of gas at a volume of .335 L if it occupies 1700 mL at 850 mm Hg?

dangina [55]

Answer:

Pressure = 4313.43mmHg

Explanation:

P1 = ?

V1 = 0.335L

V2 = 1700mL =1700*10^-3L = 1.7L

P2 = 850mmhg

From Boyle's law, the volume of a fixed mass of gas is inversely proportional to its pressure provided that temperature remains constant.

P = k / v

K = pv. P1V1 = P2V2 = P3V3 =........=PnVn

P1V1 = P2V2

Solve for P1,

P1 = (P2*V2) / V1

P1 = (850 * 1.7) / 0.335

P1 = 4313.43mmHg

The pressure of the gas was 4313.43mmHg

7 0

3 years ago

indicate whether the entropy of the system increases or decreases. Mixing 10 mL of 90.0 °C water with 10 mL of 10 °C water. The

Ivanshal [37]

Answer:

When the water is mixed with water at lower temperature the effective temperature of the system (i.e the water at lower temperature) will increase, thereby increasing it's entropy

Explanation:

The answer that "the entropy will is increases" is correct as:

The water at 90° C i.e at higher temperature is mixed with the water at 10° C i.e the water at the lower temperature.

The water at lower temperature will have molecules with lower energy while the water with higher temperature will have molecules undergoing high thermal collisions. Thereby, when the water is mixed with water at lower temperature the effective temperature of the system (i.e the water at lower temperature) will increase, thereby increasing it's entropy.

Therefore, the answer is correct with respect to the water at lower temperature.

Meanwhile, for the water at higher temperature , the temperature of the system will decrease. Thus, the entropy of the water at higher level will decrease.

5 0

3 years ago

Li+1 had gained ______ one electron

zysi [14]

Maybe molecules one electron

3 0

3 years ago

Name the types of salts.

rosijanka [135]

Answer:

that is my answer

Explanation:

the HyPO salt

4 0

2 years ago

How many moles, kmols in: 100 g of CO2, 1 litre of ethyl alcohol of density 0.789 g/cm3 and a) 1.5m3 of O2 at 25°C and 1 atm. b)

Nutka1998 [239]

Explanation:

1) Mass of carbon dioxide = 100 g

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = $\frac{100 g}{44 g/mol}=2.273 moles$

1 mol = 0.001 kmol

2.273 moles= 2.273 × 0.001 kmol = $2.273\times 10^{-3} kmol$

2) 1 liter of ethyl alcohol of density $0.789 g/cm^3$

Volume of ethyl alcohol ,V= 1 L = 1000 mL

Density of ethyl alcohol =d = $0.789 g/cm^3$

$1 cm^3=1 mL$

Mass of ethyl alcohol = m

$m=d\times V=0.789 g/cm^3\times 1000 mL=789 g$

Molar mass of ethyl alcohol = 46 g/mol

Moles of ethyl alcohol = $\frac{789 g}{46 g/mol}=17.152 mol$

$17.152 mol=17.152\times 0.001 kmol=1.7152\times 10^{-4} kmol$

3) Volume of oxygen gas,V = $1.5 m^3=1500 L$

$1 m^3= 1000 L$

Temperature of the gas = T= 25°C = 298.15 K

Pressure of the gas ,P= 1 atm

Moles of oxygen gas = n

$PV=nRT$

$n=\frac{RT}{PV}=\frac{0.0821 atm L/mol K\times 298.15 K}{1 atm\times 1500 L}=0.01632 mol$

0.01632 mol = 0.01632 × 0.001 kmol= $1.632\times 10^{-5} kmol$

4) Volume water in mixture = 1 L

Density of water = $1000 kg/m^3=\frac{1,000,000 g}{1000 L}=1000 g/L$

Mass of water = $1000 g/L\times 1 L = 1000 g$

Volume of alcohol = 2.5 L

Density of alcohol = $789 kg/m^3=\frac{789000 g}{1000 L}=789 g/L$

Mass of alcohol = $789 g/L\times 2.5 L = 1972.5 g$

Mass of mixture = 1000 g + 1972.5 g = 2972.5 g

Mass percentage of water :

$\frac{1000 g}{2972.5 g}\times 100=33.64\%$

Mass percentage of alcohol :

$\frac{1972.5 g}{2972.5 g}\times 100=66.36\%$

Moles of water :

$n_1=\frac{1000 g}{18 g/mol}=55.55 mol$

Moles of alcohol =

$n_2=\frac{1972.5 g}{46 g/mol}=42.88 mol$

Mole fraction of water :

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{55.55 mol}{55.55 mol+42.88 mol}=0.5644$

Mole fraction of alcohol :

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{42.88 mol}{55.55 mol+42.88 mol}=0.4356$

3 0

2 years ago