Answer:
1. Volume as STP = 755 L
2. Outside temperature = 255 K
3. Percentage yield = 70.5%
Explanation:
1. At STP, pressure = 101.3 kpa, temperature = 0°C or 273.15 K
Using the general gas equation :
P1V1/T1 = P2V2/T2
P1 = 620 kpa
V1 = 140 L
T1 = 37°C or (273.15 + 37) K = 310.15 K
P2 = 101.3 kpa
V2 = ?
T2 = 273.15 K
V2 = P1V1T2/P2T1
V2 = 620 × 140 × 273.15 / 101.3 × 310.15
V2 = 755 L
2. Using Charles' gas law:
V1/T1 = V2/T2
V1 = 2.5 L
T1 = 290 K
V2 = 2.2 L
T2 = ?
T2 = V2T1/VI
T2 = 2.2 × 290 / 2.5
T2 = 255 K
3. Equation of reaction : 2 Al + 3 CuSO4 ---> Al2 (SO4)3 + 3 Cu
From equation of the reaction, 2 moles of Al produces 3 moles of Cu
Molar mass of Al = 27 g; Molar mass of Cu = 63.5 g
2 moles of Al = 2 × 27 g = 54 g; 3 moles of Cu = 3× 63.5 = 190.5 g
54 g of Al produces 190.5 g of Cu
1.87 g of Al will produce 190.5/54 × 1.87 g of Cu = 6.60 g of Cu
Percentage yield = actual yield /theoretical yield × 100%
Percentage yield = 4.65/6.60 × 100%
Percentage yield = 70.5%
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.
According to Kepler's second law of orbital motion, a plane's orbital speed changes , depending on how far it is from the sun. The closer a planet is to the sun, the stronger the sun's gravitational pull on it, and the faster the planet moves. The farther away from the sun, the weaker the sun's gravitational pull and the slower it moves in its orbit.
The orbit of a planet around the sun is not a perfect circle, but an ellipse - a flattened circle.
An intensive property is a property that does not change depending on how much mass of it you are considered. An example of an intensive property is density. No matter how much water you examine, the density of the sample will be 1g/cm³.
The answer is 0.975 L
Volume = mol/Molarity
We have molarity (0.788 M) and we need mol and volume. Let's first calculate number of moles of CaCl2 in 85.3 g:
Molar mass of CaCl2 is sum of atomic masses of Ca and Cl:
Mr(CaCl2) = Ar(Ca) + 2Ar(Cl) = 40 + 2 * 35.45 = 40 + 70.9 = 110.9 g/mol
So, if 110.9 g are in 1 mol, 85.3 g will be in x mol:
110.9 g : 1 mole = 85.3 g : x
x = 85.3 g * 1 mole / 110.9
x = 0.769 moles
Now, calculate the volume:
V = 0.769/0.788
V = 0.975 L
