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IrinaVladis [17]
2 years ago
8

State whether each of the following equations represents a synthesis (s), decomposition(d), single replacement

Chemistry
1 answer:
Rzqust [24]2 years ago
6 0
Dr stay safe always share my answer
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At 700 K, the reaction2SO2(g) + O2(g) 2SO3(g)has the equilibrium constant Kc = 4.3 x 106, and the following concentrations are p
dedylja [7]

Explanation:

The given reaction is as follows.

       2SO_{2} + O_{2}(g) \rightarrow 2SO_{3}(g)

Value of equilibrium constant is given as K_{c} = 4.3 \times 10^{6}[/tex].

Concentration of given species is [SO_2] = 0.010 M; [SO_3] = 10.M; [O_2] = 0.010 M.

Formula for experimental value of equilibrium constant (Q) is as follows.

             Q = \frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}

Putting the given concentration as follows.

              Q = \frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}

             Q = \frac{(10)^{2}}{(0.010)^{2}(0.010)}

              Q = 10^{8}

It is known that when Q > K_{eq}, then reaction moves in the backward direction.

When Q < K_{eq}, then reaction moves in the forward direction.

When Q = K_{eq}, then reaction is at equilibrium.

As, for the given reaction Q > K_{eq} then it means reaction moves in the backward direction.

Thus, we can conclude that the reaction is moving in the backward direction, that is, right to left to reach the equilibrium.

5 0
3 years ago
What volume of water would be added to 16.5 ml of a 0.0813 m solution of sodium borate in order to get a 0.0200 m s?
raketka [301]
When diluting solutions from concentrated solutions the following formula can be used 
c1v1 = c2v2 
where c1 is concentration and v1 is volume of the concentrated solution 
and c2 is concentration and v2 is volume of the diluted solution to be prepared
substituting these values 
0.0813 M x 16.5 mL = 0.0200 M x V
V = 67.1 mL
the volume of the diluted solution prepared is 67.1 mL.
the volume of water that should be added to get a final volume of 67.1 mL is (67.1 - 16.5 ) = 50.6 mL
a volume fo 50.6 mL should be added 
7 0
3 years ago
What is the entropy change in the environment when 5.0 MJ of energy is transferred thermally from a reservoir at 1000 K to one a
Leni [432]

Answer:

The entropy change in the environment is 3.62x10²⁶.

Explanation:

The entropy change can be calculated using the following equation:

\Delta S = \frac{Q}{k_{B}}(\frac{1}{T_{f}} - \frac{1}{T_{i}})

Where:

Q: is the energy transferred = 5.0 MJ

k_{B}: is the Boltzmann constant = 1.38x10⁻²³ J/K  

T_{i}: is the initial temperature = 1000 K

T_{f}: is the final temperature = 500 K

Hence, the entropy change is:

\Delta S = \frac{5.0 \cdot 10^{6} J}{1.38 \cdot 10^{-23} J/K}(\frac{1}{500 K} - \frac{1}{1000 K}) = 3.62 \cdot 10^{26}                                    

Therefore, the entropy change in the environment is 3.62x10²⁶.

I hope it helps you!          

7 0
3 years ago
Explain what would happen to the air temperature at Riverdale School over the course of the
monitta

Answer:

Air temperature increases.

Explanation:

The air temperature at Riverdale School increases if the groundwater system were used because this system uses very huge amount of water than the other system. By using more molecules having the same temperature, the thing with more molecules has more total kinetic energy or thermal energy than the thing with fewer molecules so that's why air temperature of the school is increases.

8 0
2 years ago
A wafer of gold measuring 15 cm x 20 cm has a mass of 600 g. How thick is this wafer in mm ?
Gre4nikov [31]
600,000 mm if im not mistaken.
7 0
3 years ago
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