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andrew-mc [135]
3 years ago
12

Calculate the number of calories needed to increase the temperature of 50.0 g of copper metal from 21.0 degrees C to 75.0 degree

s C. Given, the specific heat of copper is 0.382 cal/g x degrees C.
Chemistry
1 answer:
KonstantinChe [14]3 years ago
4 0
<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.

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GarryVolchara [31]

Answer:

4.823 x 10^-19 J

Explanation:

Energy is calculated by E = hv where h - Planck's constant in joule.s

v - frequency.

in this particular question the wave length is 4.12 x 10^-7 m. to exhaustively use this we need a relation between wave length & frequency. c=wv where C is approximately 3 x 10^8m/s

-v = c/w = 3x10^8m/s / 4.12 x 10^-7m = 7.28 x 10^14 Hz or 1/sec

now we can simply use Planck's constant in E=hv =

(6.626 x 10^-34) x (7.28 x 10^14Hz) = 4.823 x 10^-19 J.

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2 years ago
In which of the following reactions does a decrease in the volume of the reaction vessel at constant
Black_prince [1.1K]

Answer:

The correct option is: A) 2H₂(g) + O₂(g) → 2H₂O(g)

Explanation:

According to the Le Chatelier's principle, change in the volume of the reaction system causes equilibrium to shift in the direction that reduces the effect of the volume change.

When the <u>volume decreases then the pressure of the reaction vessel increases, then the equilibrium shifts towards the reaction side that produces less number of moles of gas.</u>

<u />

A) 2H₂(g) + O₂(g) → 2H₂O(g)

The number of moles of reactant is 3 and number of moles of product is 2.

<u>Therefore, when volume decreases, the equilibrium shifts towards the product side, thereby </u><em><u>favoring the formation of products.</u></em>

B) NO₂(g) + CO(g) → NO(g) + CO₂(g)

The number of moles of reactant and product both is 2.

<u>Therefore, when the volume decreases, the equilibrium does not shift in any direction.</u>

C) H₂(g) + I₂(g) → 2HI(g)

The number of moles of reactant and product both is 2.

<u>Therefore, when the volume decreases, the equilibrium does not shift in any direction.</u>

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The number of moles of reactant is 2 and number of moles of product is 3.

<u>Therefore, when volume decreases, the equilibrium shifts towards the reactant side, thereby </u><em><u>favoring the formation of reactants.</u></em>

E) MgCO₃(s) → MgO(s) + CO₂(g)

The number of moles of reactant is 1 and number of moles of product is 2.

<u>Therefore, when volume decreases, the equilibrium shifts towards the reactant side, thereby </u><em><u>favoring the formation of reactants.</u></em>

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