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sveticcg [70]
3 years ago
5

Please help me. I can’t give too much points bc I don’t have too much.

Chemistry
1 answer:
zubka84 [21]3 years ago
7 0
Answer is photons hope this helps
You might be interested in
The goal of this lesson:
mote1985 [20]

Answer:

1. The concentration of N₂O₄ decrease as the temperature of the system increased.

2. The formation of products was favored by the addition of heat.

3. The reaction going from right to left is exothermic.

4. N₂O₄ → 2NO₂; HR = +14 kcal.

Explanation:

Equation of the reaction is as follows: N₂O₄ ⇄ 2NO₂

The forward reaction proceeds with an increase in temperature. As the temperature of the reaction was increased, more of the N₂O₄ decomposed to form NO₂. Therefore, the concentration of N₂O₄ deceased.

2. The formation of products, that is the forward reaction leading to formation of NO₂ was favored by the addition of heat. Thus, with increase in temperature, the concentration of NO₂ increased.

3. An endothermic reaction is a type of reaction which requires energy input in the form of heat in order to proceed.

From the observations during the reaction, an increase in temperature by the addition of heat resulted in an increase in the forward reaction, therefore, the forward reaction is endothermic and the backward reaction is exothermic. Thus, the reaction in which colorless N₂O₄ is produced, is an exothermic reaction.

4. The change in enthalpy of a reaction is the difference in the heat content of reactants and products. For exothermic reactions, enthalpy change is negative, whereas for endothermic reactions, enthalpy change is positive.

The decomposition of  N₂O₄ to NO₂ is an endothermic reaction. Hence, the correct chemical equation is: N₂O₄ → 2NO₂; HR = +14 kcal.

7 0
3 years ago
I want to know the steps.
Artyom0805 [142]

The answer for the following problem is described below.

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

Explanation:

Given:

enthalpy of combustion of glucose(ΔH_{f} of C_{6}H_{12} O_{6}) =-1275.0

enthalpy of combustion of oxygen(ΔH_{f} of O_{2}) = zero

enthalpy of combustion of carbon dioxide(ΔH_{f} of CO_{2}) = -393.5

enthalpy of combustion of water(ΔH_{f} of H_{2} O) = -285.8

To solve :

standard enthalpy of combustion

We know;

ΔH_{f}  = ∈ΔH_{f} (products) - ∈ΔH_{f} (reactants)

C_{6}H_{12} O_{6} (s) +6 O_{2}(g) → 6 CO_{2} (g)+ 6 H_{2} O(l)

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [0 - 1275]

ΔH_{f} = 6 (-393.5) + 6(-285.8)  - 0 + 1275

ΔH_{f} = -2361 - 1714 - 0 + 1275

ΔH_{f} =-2800 kJ

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

7 0
3 years ago
Which activity might help to increase the validity of this experiment?
MArishka [77]
A should be the answer because the more you test an experiment the more data you have to rely on changing the experiment would cause you to have different outcomes making the results different and unreliable so B, C, and D is not going to be the answer Hope this helps
5 0
3 years ago
9.
Sholpan [36]

Answer: Your mass would be the same, but you would weigh less because the gravity on the moon is less than it is on Earth.

Explanation:

7 0
3 years ago
A reaction produces 92.50 g FeSO4. How many grams of CuSO4 are necessary for this to occur?
xxMikexx [17]

Fe+CuSO4⟶Cu+FeSO4

Given that  

FeSO4 = 92.50 g  

Number of moles = amount in  g / molar mass

=92.50 g / 151.908 g/mol

=0.609 moles FeSO4

Now calculate the moles of CuSO4 as follows:

0.609 moles FeSO4 * 1 mole CuSO4 /1 mole FeSO4

= 0.609 moles CuSO4

Amount in g = number of moles * molar mass

= 0.609 moles CuSO4 * 159.609 g/mol

= 97.19 g CuSO4


3 0
3 years ago
Read 2 more answers
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