All categories

3 years ago

In terms of energy, how would you classify the following chemical reaction? 2CuO + 315 kJ → 2Cu + O2

2 answers:

6 0

"Endothermic" is the way among the following choices given that you would classify the following chemical reaction, in terms of energy. The correct option among all the options that are given in the question is the second option or option "B". I hope that the answer has come to your help.

omeli [17]3 years ago

6 0

Answer: The reaction 2CuO + 315 kJ → 2Cu + O2 is an endothermic reaction. (Option B)

Explanation: In order to determine what type of reaction is occurring above, you need to know that if heat is being released then it is an exothermic reaction. It can be determined if it would be on the product side. Whilst, if the heat is being absorbed, it would be an endothermic reaction and it would be on the side of the reactants.

You might be interested in

HELP!!!

jek_recluse [69]

Answer:

Geothermal energy has a faster rate because a vast amount of heat energy is always present inside Earth's core.

Geothermal energy has a slower rate because the water present below Earth's surface cools down periodically

Explanation:

6 0

3 years ago

Read 2 more answers

There are two main items that cause the seasons what are they. There more than one answer

V125BC [204]

What causes the seasons?
The seasons are caused by the tilt of the Earth's rotational axis away or toward the sun as it travels through its year-long path around the sun.

8 0

3 years ago

If Kp = 0.15, which statement is true of the reaction mixture before any reaction occurs?

Fiesta28 [93]

The given sentence is part of a longer question.

I found this question with the same sentence. So, I will help you using this question:

For the reaction N2O4(g) ⇄ 2NO2(g), a reaction mixture at a certain temperature initially contains both N2O4 and NO2 in their standard states (meaning they are gases with a pressure of 1 atm). If Kp = 0.15, which statement is true of the reaction mixture before any reaction occurs?

(a) Q = K; The reaction is at equilibrium.
(b) Q < K; The reaction will proceed to the right.
(c) Q > K; The reaction will proceed to the left.

The answer is the option (c) Q > K; The reaction will proceed to the left, since Qp = 1, and 1 > 0.15.
Explanation:

Kp is the equilibrium constant in term of the partial pressures of the gases.

Q is the reaction quotient. It is a measure of the progress of a chemical reaction.

The reaction quotient has the same form of the equilibrium constant but using the concentrations or partial pressures at any moment.

At equilibrium both Kp and Q are equal. Q = Kp

If Q < Kp then the reaction will go to the right (forward reaction) trying to reach the equilibrium,

If Q > Kp then the reaction will go to the left (reverse reaction) trying to reach the equilibrium.

Here, the state is that both pressures are 1 atm, so Q = (1)^2 / 1 = 1.

Since, Q = 1 and Kp = 0.15, Q > Kp and the reaction will proceed to the left.

6 0

3 years ago

In the following reaction, how many grams of NaBr will produce 244 grams of NaNO3? Pb(NO3)2(aq) 2 NaBr(aq) PbBr2(s) 2 NaNO3(aq)

lyudmila [28]

Answer : The mass of $NaBr$ is, 295.323 grams

Solution :

First we have to calculate the moles of $NaNO_3$ .

$\text{Moles of }NaNO_3=\frac{\text{Mass of }NaNO_3}{\text{Molar mass of }NaNO_3}=\frac{244g}{85g/mole}=2.87moles$

Now we have to calculate the moles of NaBr.

The balanced chemical reaction is,

$Pb(NO_3)_2(aq)+2NaBr(aq)\rightarrow PbBr_2(s)+2NaNO_3(aq)$

From the balanced reaction we conclude that

As, 2 moles of $NaNO_3$ react with 2 moles of $NaBr$

So, 2.87 moles of $NaNO_3$ react with 2.87 moles of $NaBr$

Now we have to calculate the mass of NaBr.

$\text{Mass of }NaBr=\text{Moles of }NaBr\times \text{Molar mass of }NaBr$

$\text{Mass of }NaBr=(2.87mole)\times (102.9g/mole)=295.323g$

Therefore, the mass mass of $NaBr$ is, 295.323 grams

7 0

3 years ago

Silver occurs in trace amounts in some ores of lead, and lead can displace silver from solution: Pb(s) + 2Ag+ (aq) LaTeX: \longr

VikaD [51]

Answer : The value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

Explanation :

The balanced cell reaction will be,

$Pb(s)+2Ag^+(aq)\rightarrow Pb^{2+}(aq)+2Ag(g)$

The half-cell reactions are:

Oxidation reaction (anode) : $Pb(s)\rightarrow Pb^{2+}(aq)+2e^-$

Reduction reaction (cathode) : $2Ag^+(aq)+2e^-\rightarrow 2Ag(g)$

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

where,

$\Delta G^o$ = standard Gibbs free energy

F = Faraday constant = 96500 C

n = number of electrons in oxidation-reduction reaction = 2

$E^o_{cell}$ = standard electrode potential of the cell = 0.93 V

Now put all the given values in the above formula, we get:

$\Delta G^o=-2\times 96500\times 0.93$

$\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol$

Now we have to calculate the value of 'K'.

$\Delta G^o=-RT\ln K$

where,

$\Delta G_^o$ = standard Gibbs free energy = -180 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

$-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K$

$K=3.6\times 10^{31}$

Therefore, the value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

5 0

3 years ago