All categories

2 years ago

Calculate the molar mass of cacl2

Chemistry

1 answer:

kirza4 [7]2 years ago

6 0

Answer:

$\boxed {\boxed {\sf 110.98 \ g/mol}}$

Explanation:

The molar mass is the mass of a substance in grams per mole.

To find it, add the mass of each element in the compound. These masses can be found on the Periodic Table.

The compound given is:

$CaCl_2$

The compound has 1 Ca (calcium) and 2 Cl (chlorine).

Mass of Calcium

The molar mass of calcium is 40.08 g/mol
There is only one atom of Calcium in CaCl₂, so the number above is what we will use.

Mass of Chlorine

The molar mass of chlorine is 35.45 g/mol
There are two atoms of chlorine in CaCl₂, therefore we need to multiply the molar mass by 2.
35.45 * 2= 70.9 g/mol

Molar Mass of CaCl₂

Now, to find the molar mass, add the molar mass of 1 calcium and 2 chlorine.
40.08 g/mol + 70.9 g/mol =110.98 g/mol

The molar mass of CaCl₂ is 110.98 grams per mole.

You might be interested in

What is the ph of a solution with h+ = 7.0* 10

Mekhanik [1.2K]

Do you mean h=7.0+10? If so your answer is 70.

8 0

2 years ago

How many grams of mg are in 534g of mgo

Kitty [74]

Answer:

0.534mg..

....

...

....

7 0

2 years ago

. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are

olga nikolaevna [1]

Answer: The standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

Explanation:

We are given:

$K_{sp}\text{ of }Ag_2S=8\times 10^{-51}$

Relation between standard Gibbs free energy and equilibrium constant follows:

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

where,

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

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

K = equilibrium constant or solubility product = $8\times 10^{-51}$

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ$

For the given chemical equation:

$Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)$

The equation used to calculate Gibbs free change is of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

The equation for the Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]$

We are given:

$\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ$

Putting values in above equation, we get:

$285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol$

Hence, the standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

8 0

3 years ago

The entropy change for a real, irreversible process is equal to:______.

taurus [48]

Answer:

The entropy change for a real, irreversible process is equal to zero.

The correct option is 'c'.

Explanation:

Lets look around all the given options -:

(a) the entropy change for a theoretical reversible process with the same initial and final states , since the entropy change is equal and opposite in reversible process , thus this option in not correct.

(b) equal to the entropy change for the same process performed reversibly ONLY if the process can be reversed at all. Since , the change is same as well as opposite too . Therefore , this statement is also not true .

(c) zero. This option is true because We generate more entropy in an irreversible process. Because no heat moves into or out of the surroundings during the procedure, the entropy change of the surroundings is zero.

(d) impossible to tell. This option is invalid , thus incorrect .

Hence , the correct option is 'c' that is zero.

8 0

2 years ago

Name the acids: HBrO

solong [7]

Answer:

Hypobromous acid

Explanation:

6 0

3 years ago