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Nezavi [6.7K]
3 years ago
11

A 12.0% sucrose solution by mass has a density of 1.05 gem, what mass of sucrose is present in a 32.0-mL sample of this solution

?
A) 0.126g
B) 3.66g
C) 4.03g
D) 3.84g
E) 280 g
Chemistry
1 answer:
natulia [17]3 years ago
7 0

Answer:

Option C. 4.03 g

Explanation:

Firstly we analyse data.

12 % by mass, is a sort of concentration. It indicates that in 100 g of SOLUTION, we have 12 g of SOLUTE.

Density is the data that indicates grams of solution in volume of solution.

We need to determine, the volume of solution for the concentration

Density = mass / volume

1.05 g/mL = 100 g / volume

Volume =  100 g / 1.05 g/mL → 95.24 mL

Therefore our 12 g of solute are contained in 95.24 mL

Let's finish this by a rule of three.

95.24 mL contain 12 g of sucrose

Our sample of 32 mL may contain ( 32 . 12) / 95.24 = 4.03 g

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Explanation:

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A gas has a volume of 4.25 m3 at a temperature of 95.0°C and a pressure of 1.05 atm. What temperature will the gas have at a pre
Goryan [66]

Answer:

\boxed {\boxed {\sf 82.7 \textdegree C}}

Explanation:

We are asked to find the temperature of a gas given a change in pressure and volume. We will use the Combined Gas Law, which combines 3 gas laws: Boyle's, Charles's, and Gay-Lussac's.

\frac {P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

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\frac {1.05 \ atm * 4.25 \ m^3}{95.0 \textdegree C}= \frac{P_2V_2}{T_2}

Then, the pressure increases to 1.58 atmospheres and the volume decreases to 2.46 cubic meters.

\frac {1.05 \ atm * 4.25 \ m^3}{95.0 \textdegree C}= \frac{1.58  \ atm *2.46 \ m^3}{T_2}

We are solving for the new temperature, so we must isolate the variable T₂. Cross multiply. Multiply the first numerator by the second denominator, then multiply the first denominator by the second numerator.

(1.05 \ atm * 4.25 \ m^3) * T_2 = (95.0 \textdegree C)*(1.58 \ atm * 2.46 \ m^3)

Now the variable is being multiplied by (1.05 atm * 4.25 m³). The inverse operation of multiplication is division, so we divide both sides by this value.

\frac {(1.05 \ atm * 4.25 \ m^3) * T_2}{(1.05 \ atm * 4.25 \ m^3)} = \frac{(95.0 \textdegree C)*(1.58 \ atm * 2.46 \ m^3)}{(1.05 \ atm * 4.25 \ m^3)}

T_2=\frac{(95.0 \textdegree C)*(1.58 \ atm * 2.46 \ m^3)}{(1.05 \ atm * 4.25 \ m^3)}

The units of atmospheres and cubic meters cancel.

T_2=\frac{(95.0 \textdegree C)*(1.58* 2.46 )}{(1.05 * 4.25 )}

Solve inside the parentheses.

T_2= \frac{(95.0 \textdegree C)*3.8868}{4.4625}

T_2= \frac{369.246}{4.4625} \textdegree C}

T_2 = 82.74420168 \textdegree C

The original values of volume, temperature, and pressure all have 3 significant figures, so our answer must have the same. For the number we calculated, that is the tenths place. The 4 in the hundredth place to the right tells us to leave the 7 in the tenths place.

T_2 \approx 82.7 \textdegree C

The temperature is approximately <u>82.7 degrees Celsius.</u>

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