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3 years ago

12 g of powdered magnesium oxide reacts with nitric acid to

Chemistry

1 answer:

galben [10]3 years ago

4 0

Answer:

80.8 g

Explanation:

First, let's write a balanced equation of this reaction

MgO + 2HNO₃ → Mg(NO₃)₂ + H₂O

Now let's convert grams to moles

We gotta find the weight of MgO

24 + 16 = 40 g/mol

12/40 = 0.3 moles of MgO

We can use this to find out how much Magnesium Nitrate will be formed

0.3 x 1 MgO / 1 Mg(NO₃)₂ = 0.3 moles of Magnesium Nitrate formed

Convert moles to grams

Find the weight of Mg(NO₃)₂ but don't forget that 2 subscript acts as a multiplier of whatever is inside that parenthesis.

24 + 14 x 2 + 16 x 3 x 2 = 148 g/mol

148 x 0.3 = 80.8 g

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Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it

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Explanation:

We are given:

Initial concentration of $COF_2$ = 2.00 M

The given chemical equation follows:

$2COF_2(g)\rightleftharpoons CO_2(g)+CF_4(g)$

Initial: 2.00

At eqllm: 2.00-2x x x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}$

We are given:

$K_c=6.30$

Putting values in above expression, we get:

$6.30=\frac{x\times x}{(2.00-2x)^2}\\\\x=0.834,1.25$

Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible

So, equilibrium concentration of $COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M$

Hence, the equilibrium concentration of $COF_2$ is 0.332 M

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3 years ago

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PLEASE HELP!!!!! ASAP this is due tomorrow and I don't get it help me with some problems I'll be so thankful.

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