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Agata [3.3K]
3 years ago
9

What is the molecular formula for a compound with an Empirical formula of C2H3, if the true Molar Mass is 81 g/mol:

Chemistry
1 answer:
GREYUIT [131]3 years ago
7 0

Answer:

To be honest I’d don’t even know

Explanation:

Because I’d said so

You might be interested in
Which provides a baseline for an experimental investigation
Goryan [66]

Answer:

control group

Explanation:

7 0
3 years ago
Consider this equilibrium reaction between carbon monoxide and hydrogen gas, occurring in a sealed flexible container. CO(g) + 3
wariber [46]

Answer:

More H2(g) is added to the container : <u>Towards products.</u>

CO is removed from the container : <u>Towards reactants.</u>

More CH4(g) is added to the container : <u>Towards reactants</u>

H2O(g) is removed from the container <u>: Towards products.</u>

The contents of the container are heated up. :<u> Towards the reactants.</u>

The contents of the container are cooled down : <u>Towards the products.</u>

The pressure inside the container is increased. :<u>Towards the products</u>

The container is stretched to increase the volume: <u>Towards the reactants.</u>

Explanation: :

CO(g) + 3 H2g) → CH4(g) + H2O(g)+ heat

There is released heat, so this reaction is exothermic

If the H2 concentration is increased, the system will try to change the concentration change by shifting the balance to the right, and thus the concentration of products will increase.<u> Towards products.</u>

If the CO is removed, the system will try to change this situation by shifting the balance to the left, and thus the concentration of reactants will increase, the concentration of products will decrease. <u>Towards reactants.</u>

If the CH4 concentration is increased, the system will try to change the concentration change by shifting the balance to the left, and thus the concentration of reactants will increase. <u>Towards reactants</u>

If the H2O is removed, the system will try to change this situation by shifting the balance to the right, and thus the concentration of products will increase, the concentration of products will decrease. <u>Towards products.</u>

If the temperature is increased, the system will reduce the amount of heat released. So the balance will shift to the left. <u>Towards the reactants.</u>

This because the extra heat / energy must be used.

If the temperature is decreased, the system will produce more heat  So the balance will shift to the right. <u>Towards the products.</u>

This because more heat /energy needs to be produced to make up for the loss of heat (energy).

If the pressure is increased, the system will shift to the side with fewer moles of gas. In this case, there are 4 moles on the left and 2 moles on the right.  So the balance will shift to the right. <u>Towards the products.</u> An increase of pressure has the same effect on the equilibrium as a decrease of the volume.

If the volume is increased, this means the pressure is decreased, the system will shift to the side with most moles of gas. In this case, there are 4 moles on the left and 2 moles on the right.  So the balance will shift to the left. <u>Towards the reactants.</u> An increase of volume has the same effect on the equilibrium as a decrease of the pressure.

6 0
3 years ago
The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +
Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b) The solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

Explanation:

(a)  Chemical equation for the given reaction in pure water is as follows.

           CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)

Initial:                         0            0

Change:                    +x           +x

Equilibm:                   x             x

K_{sp} = 1.2 \times 10^{-6}

And, equilibrium expression is as follows.

          K_{sp} = [Cu^{+}][Cl^{-}]

       1.2 \times 10^{-6} = x \times x

             x = 1.1 \times 10^{-3} M

Hence, the solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b)  When NaCl is 0.1 M,

       CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq),  K_{sp} = 1.2 \times 10^{-6}

   Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq),  K = 8.7 \times 10^{4}

Net equation: CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

               K' = K_{sp} \times K

                          = 0.1044

So for, CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

Initial:                     0.1                 0

Change:                -x                   +x

Equilibm:            0.1 - x                x

Now, the equilibrium expression is as follows.

              K' = \frac{CuCl_{2}}{Cl^{-}}

         0.1044 = \frac{x}{0.1 - x}

              x = 9.5 \times 10^{-3} M

Therefore, the solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

7 0
3 years ago
Pls help me out with this question!
jeka94
At STP, copper (Cu) would be the only substance here that will exist in the solid state.
4 0
2 years ago
Side Show
den301095 [7]

Answer: if you have this app then you’r not smart

Explanation: because cheating is not allowed

4 0
3 years ago
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