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3 years ago

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What is the oxidation state of each element in the compound CaS04? Include +or - in your answer as appropriate

1 answer:

borishaifa [10]3 years ago

5 0

Answer:therefore the oxidation state of sulphur is +6 in the compound CaSO4.

Explanation:

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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

Matter is more likely to exist in the ________ state as the temperature is lowered.

katovenus [111]

As the temperature is lowered matter is more likely to exist in the solid state

7 0

3 years ago

The edge length of the unit cell of KCl (NaCl-like structure, FCC) is 6.28 Å. Assuming anion-cation contact along the cell edge,

Mila [183]

Answer:

1.33 Å

Explanation:

Given that the edge length , a of the KCl which forms the FCC lattice = 6.28 Å

Also,

For the FCC lattice in which the anion-cation contact along the cell edge , the ratio of the radius of the cation to that of anion is 0.731.

Thus,

$\frac {r^+}{r^-}=0.731$ .................1

Also, the sum of the radius of the cation and the anion in FCC is equal to half of the edge length.

Thus,

$r^++r^-=\frac {a}{2}$ ...................2

Given that:

$Cl^-\ (r^-) = 1.82\ \dot{A}$

To find,

$K^+\ (r^+) = ? \dot{A}$

Using 1 and 2 , we get:

$1.731\ r^+=0.731\times \frac {6.28}{2}$

<u>Size of the potassium ion = 1.33 Å</u>

4 0

3 years ago

1 webbed feet <br> 2wing structure <br> 3beaks <br> 4 eye color

malfutka [58]

I believe it's webbed feet

7 0

3 years ago

Read 2 more answers

What is the difference between speed and acceleration?<br> Please help :)

ioda

Answer:

Speed is how fast you are going at the moment but acceleration is how fast you are building up speed to get to a set point

Explanation:

4 0

3 years ago