Chemical reactions often involve changes in energy due to the breaking and formation of bonds. Reactions in which energy is released are exothermic reactions, while those that take in heat energy are endothermic.

3 0

3 years ago

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.

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

Initially, the gas has a pressure of 1.05 atmospheres, a volume of 4.25 cubic meters, and a temperature of 95.0 degrees Celsius.

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

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>

3 0

3 years ago

Dominic and Eva are using the same type of stopwatches to measure the time it takes for a chemical reaction to occur. They each

SIZIF [17.4K]

They did not repeat their tests multiple times.

Reason being is using the same stopwatch doesn’t matter and changing the reactants won’t help, and you without a doubt need more than one person collecting the data so,

They did not repeat their tests multiple times is the answer.

4 0

1 year ago

Identify all statements that accurately describe the structure of the pyruvate dehydrogenase ( PDH ) complex.

Nostrana [21]

Answer:

The statements that correctly describes pyruvate dehydrogenase includes:

- Several copies each of E 1 and E 3 surround E 2.

-A regulatory kinase and phosphatase are part of the mammalian PDH complex.

-E 2 contains three domains.

Explanation:

Pyruvate dehydrogenase is a hydrolase key enzyme in glucose metabolism which converts pyruvate to acetyl- ChoA. It also forms a complex that catalyzes an irreversible reaction that is the entry point of pyruvate into the TCA cycle. Pyruvate dehydrogenase complex contains E1, E2 and E3 enzymes that transform pyruvate, NAD+, coenzyme A into acetyl-CoA, CO2, and NADH. Also, A regulatory kinase and phosphatase are part of the mammalian PDH complex and E 2 contains three domains.

5 0

3 years ago