When fossil fuels are burned, _____ is released into the atmosphere.?

attashe74 [19]

During Photosynthesis , Plants Transforms Light Energy Into Chemical Energy .

6 0

3 years ago

Carbon tetrachloride can be produced by the following reaction: Suppose 1.20 mol of and 3.60 mol of were placed in a 1.00-L flas

hjlf

The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Suppose 1.20 mol $CS_2(g)$ of and 3.60 mol of $Cl_2(g)$ were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol of $CCl_4$ . Calculate equilibrium constant at the unknown temperature.

Answer: The equilibrium constant at unknown temperature is 0.36

Explanation:

Moles of $CS_2$ = 1.20 mole

Moles of $Cl_2$ = 3.60 mole

Volume of solution = 1.00 L

Initial concentration of $CS_2$ = $\frac{moles}{volume}=\frac{1.20mol}{1L}=1.20M$

Initial concentration of $Cl_2$ = $\frac{moles}{volume}=\frac{3.60mol}{1L}=3.60M$

The given balanced equilibrium reaction is,

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Initial conc. 1.20 M 3.60 M 0 0

At eqm. conc. (1.20-x) M (3.60-3x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}$

Now put all the given values in this expression, we get :

$K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}$

Given :Equilibrium concentration of $CCl_4$ , x = $\frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M$

$K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}$

$K_c=0.36$

Thus equilibrium constant at unknown temperature is 0.36

4 0

3 years ago

PLEEEASSEEEE HELPPPPP!!!!!Take some time to research a utility plant. If there is one in your area, you may even visit it.

raketka [301]

Answer:

There are four laws of thermodynamics that define fundamental physical quantities (temperature, energy, and entropy) and that characterize thermodynamic systems at thermal equilibrium.

Explanation:

6 0

3 years ago

Consider two gases, A and B, are in a container at room temperature. What effect will the following changes have on the rate of

Oksanka [162]

Answer:

B: increase.

Explanation:

When we are considering two gases A and B in a container at room temperature .

We have to find the change on rate of reaction when the number of molecules of gases A is doubled

Let [A]=a and [B]=b

A+B $\rightarrow$ product

Rate of reaction

$R_1=k[A][B]=kab$

We know that concentration is increases with increase in number of moles

When the number of molecules of gases A is doubled then concentration of gases A increases.

Therefore ,[A]=2a

Rate of reaction

$R_2=k(2a)(b)=2kab$

$R_2=2R_1$

Hence, the rate of reaction is 2 times the initial rate of reaction.Therefore, the rate of reaction will increase when the number of molecules of gases A is doubled.

Answer: B: increase.

4 0

4 years ago

In the diagram below, west is to the left, east is to the right, and the observer is looking toward the north. What is the direc

Damm [24]

I think the answer is c since the it’s looking at NW, the wind is coming from SE

7 0

2 years ago

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