<h3>
Answer:</h3>
1031.4 Calories.
<h3>
Explanation:</h3>
We are given;
Mass of the copper metal = 50.0 g
Initial temperature = 21.0 °C
Final temperature, = 75°C
Change in temperature = 54°C
Specific heat capacity of copper = 0.382 Cal/g°C
We are required to calculate the amount of heat in calories required to raise the temperature of the copper metal;
Quantity of heat is given by the formula,
Q = Mass × specific heat capacity × change in temperature
= 50.0 g × 0.382 Cal/g°C × 54 °C
= 1031.4 Calories
Thus, the amount of heat energy required is 1031.4 Calories.
Answer:
The farther away the planet the slower the revolution around the earth. the closer the faster.
Explanation:
its like a tetherball pole when it wraps around it gets closer and spins faster and faster untill it stops. Brainliest?
Answer:
Explanation:
The volume and amount are constant, so we can use Gay-Lussac’s Law:
At constant volume, the pressure exerted by a gas is directly proportional to its temperature.
Data:
p₁ = 1520 Torr; T₁ = 27 °C
p₂ = ?; T₂ = 150 °C
Calculations:
(a) Convert the temperatures to kelvins
T₁ = ( 27 + 273.15) K = 300.15 K
T₂ = (150 + 273.15) K = 423.15 K
(b) Calculate the new pressure
(c) Convert the pressure to atmospheres
We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.
<span>glucose-1-phosphate⟶glucose-6-phosphate ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate ΔG∘=−1.67 kJ/mol
</span>
Therefore, the reaction is a two-step process wherein glucose-6-phosphate is the intermediate product.
glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate
In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.
ΔG°,total = −7.28 kJ/mol + 1.67 kJ/mol = -5.61 kJ/mol
Then, the equation to relate ΔG° to the equilibrium constant K is
ΔG° = -RTlnK, where R is the gas constant equal to 0.008317 kJ/mol-K.
-5.61 kJ./mol = -(0.008317 kJ/mol-K)(298 K)(lnK)
lnK = 2.2635
K = e^2.2635
K = 9.62
