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2 years ago

What's the Molar Mass of O2? O 16.0 g/mole o 32.0 g/mole 1.0 g/mole O 2.0 g/mole

Chemistry

2 answers:

stiks02 [169]2 years ago

5 0

............:.................................

Elza [17]2 years ago

5 0

32.0 g/mol! Take the atomic mass off of the periodic table (16) and multiply by 2 since there are two oxygen atoms. So your answer is 32.0 g/mol.

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When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr

Amanda [17]

Answer: The value of $K_c$ for the given reaction is 1.435

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of $N_2O_4$ = 9.2 g

Molar mass of $N_2O_4$ = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M$

For the given chemical equation:

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initial: 0.20

At eqllm: 0.20-x 2x

We are given:

Equilibrium concentration of $N_2O_4$ = 0.057

Evaluating the value of 'x'

$\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$[NO_2]_{eq}=2x=(2\times 0.143)=0.286M$

$[N_2O_4]_{eq}=0.057M$

Putting values in above expression, we get:

$K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435$

Hence, the value of $K_c$ for the given reaction is 1.435

6 0

3 years ago

Following reactions FeSO4+KOH

prohojiy [21]

Explanation:

$FeSO _{4} + \: KOH \: →K _{2}SO _{4} \: + \: Fe(OH) _{2}$

5 0

2 years ago

What is the final concentration of cl- ion when 250 ml of 0.20 m cacl2 solution is mixed with 250 ml of 0.40 m kcl solution? (as

serious [3.7K]

CaCl2 and KCl are both salts which dissociate in water when dissolved. Assuming that the dissolution of the two salts are 100 percent, the half reactions are:

CaCl2 ---> Ca2+ + 2 Cl-

KCl ---> K+ + Cl-

Therefore the total Cl- ion concentration would be coming from both salts. First, we calculate the Cl- from each salt by using stoichiometric ratio:

Cl- from CaCl2 = (0.2 moles CaCl2/ L) (0.25 L) (2 moles Cl / 1 mole CaCl2)

Cl- from CaCl2 = 0.1 moles

Cl- from KCl = (0.4 moles KCl/ L) (0.25 L) (1 mole Cl / 1 mole KCl)

Cl- from KCl = 0.1 moles

Therefore the final concentration of Cl- in the solution mixture is:

Cl- = (0.1 moles + 0.1 moles) / (0.25 L + 0.25 L)

Cl- = 0.2 moles / 0.5 moles

Cl- = 0.4 moles (ANSWER)

6 0

2 years ago

HELPPP yall please help my sister won't help me

ira [324]

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$\text{Pb(OH)}_2 +2\text{HCl} \longrightarrow \text{PbCl}_2 +2 \text{H}_2 \text{O}$

$2\text{Mn} + 6\text{HI} \longrightarrow 3 \text{H}_2+2\text{MnI}_3$

3 0

2 years ago

how many grams of calcium chloride can be prepared from 60.4 G of calcium oxide and 69.0 G of hydrochloric acid and a double dis

Ksenya-84 [330]

Balanced equation:
CaO + 2 HCl --> CaCl2 + H2O 
Calculate moles of each reactant: 
60.4 g CaO / 56.08 g/mol = 1.08 mol CaO 
69.0 g HCl / 36.46 g/mol = 1.89 mol HCl 

Identify the limiting reactant: 
Moles CaO needed to react with all HCl: 
1.89 mol HCl X (1 mol CaO / 2 mol HCl) = 0.946 mol CaO 
Because you have more CaO than that available, HCl is the limiting reactant. 

Calculate moles and mass CaCl2: 
1.89 mol HCl X (1 mol CaCl2 / 2mol HCl) X 111.0 g/mol = 105 g CaCl2

8 0

2 years ago