Molar mass (easy but I’m lazy)

How many molecules of CaCl2 are equivalent to 75.9 g CaCl2

sergij07 [2.7K]

First, you need to find:
One mole of $CaCl_{2}$ is equivalent to how many grams?

Well, for this you have to look up the periodic table. According to the periodic table:
The atomic mass of Calcium Ca = 40.078 g (See in group 2)
The atomic mass of Chlorine Cl = 35.45 g (See in group 17)

As there are two atoms of Chlorine present in $CaCl_{2}$ , therefore, the atomic mass of $CaCl_{2}$ would be:

Atomic mass of $CaCl_{2}$ = Atomic mass of Ca + 2 * Atomic mass of Cl

Atomic mass of $CaCl_{2}$ = 40.078 + 2 * 35.45 = 110.978 g

Now,

110.978 g of $CaCl_{2}$ = 1 mole.
75.9 g of $CaCl_{2}$ = $\frac{75.9}{110.978} moles$ = 0.6839 moles.

Hence,
The total number of moles in 75.9g of $CaCl_{2}$ = 0.6839 moles

According to Avogadro's number,
1 mole = 1 * $6.022 * 10^{23}$ molecules
0.6839 moles = 0.6839 * $6.022 * 10^{23}$ molecules = $4.118*10^{23}$ molecules

Ans: Number of molecules in 75.9g of $CaCl_{2}$ = $4.118*10^{23}$ molecules

-i

3 0

3 years ago

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Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M 3 O 4 ( s ) − ⇀ ↽ − 3 M ( s )

Fiesta28 [93]

This is an incomplete question, here is a complete question.

Consider the decomposition of a metal oxide to its elements, where M represents a generic metal.

$M_3O_4(s)\rightleftharpoons 3M(s)+2O_2(g)$

Substance ΔG°f (kJ/mol)

M₃O₄ -9.50

M(s) 0

O₂(g) 0

What is the standard change in Gibbs energy for the reaction, as written, in the forward direction? delta G°rxn = kJ / mol.

What is the equilibrium constant of this reaction, as written, in the forward direction at 298 K?

What is the equilibrium pressure of O₂(g) over M(s) at 298 K?

Answer :

The Gibbs energy of reaction is, 9.50 kJ/mol

The equilibrium constant of this reaction is, 0.0216

The equilibrium pressure of O₂(g) is, 0.147 atm

Explanation :

The given chemical reaction is:

$PCl_3(l)\rightarrow PCl_3(g)$

First we have to calculate the Gibbs energy of reaction $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{M(s)}\times \Delta G^0_{(M(s))}+n_{O_2(g)}\times \Delta G^0_{(O_2(g))}]-[n_{M_3O_4(s)}\times \Delta G^0_{(M_3O_4(s))}]$

where,

$\Delta G^o$ = Gibbs energy of reaction = ?

n = number of moles

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

$\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (0kJ/mol)]-[1mole\times (-9.50kJ/K.mol)]$

$\Delta G^o=9.50kJ/mol$

The Gibbs energy of reaction is, 9.50 kJ/mol

Now we have to calculate the equilibrium constant of this reaction.

The relation between the equilibrium constant and standard Gibbs free energy is:

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

where,

$\Delta G^o$ = standard Gibbs free energy = 9.50kJ/mol = 9500 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = 298 K

K = equilibrium constant = ?

$9500J/mol=-(8.314J/K.mol)\times (298K)\times \ln (K)$

$K=0.0216$

The equilibrium constant of this reaction is, 0.0216

Now we have to calculate the equilibrium pressure of O₂(g).

The expression of equilibrium constant is:

$K=(P_{O_2})^2$

$0.0216=(P_{O_2})^2$

$P_{O_2}=0.147atm$

The equilibrium pressure of O₂(g) is, 0.147 atm

5 0

4 years ago

What is the definition of outer layer? *

Nataly [62]

Answer:

1 being or located on the outside; external

2 further from the middle or central part

Explanation:

4 0

3 years ago

When magnesium-28 undergoes beta decay, what is the identity and mass number of the nucleus formed?

Ugo [173]

Beta decay increases the atomic number by 1, and does not change the mass number. 28Mg12 → 0e-1 + 28Al13

3 0

3 years ago

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Question 2

GaryK [48]

O2 mol O2 2 mil CO2 O1molO2

7 0

3 years ago