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2 years ago

Please solve for Cu,N,and O an dif you can show how cause ik confused

Chemistry

1 answer:

belka [17]2 years ago

4 0

Cu⇒ 1 atom

N⇒2 atoms

O⇒6 atoms

Total 9 atoms

<h3>Further explanation </h3>

The empirical formula is the smallest comparison of atoms of compound forming elements.

A molecular formula is a formula that shows the number of atomic elements that make up a compound.

<em>(empirical formula) n = molecular formula </em>

Chemical formula : Cu(NO₃)₂

Number of Cu : 1

Number of N : 2

Number of O = 2 x 3 = 6

Total atoms in Cu(NO₃)₂ : 1 + 2 + 6 = 9

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Hey please answer this thanks.

Aleks04 [339]

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2 years ago

The equilibrium constant, Kc, for the following reaction is 0.967 at 650 K. 2NH3(g) N2(g) 3H2(g) When a sufficiently large sampl

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Answer: Concentration of $NH_3$ in the equilibrium mixture is 0.31 M

Explanation:

Equilibrium concentration of $H_2$ = 0.729 M

The given balanced equilibrium reaction is,

$2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$

Initial conc. x 0 0

At eqm. conc. (x-2y) M (y) M (3y) M

The expression for equilibrium constant for this reaction will be:

3y = 0.729 M

y = 0.243 M

$K_c=\frac{[y]\times [3y]^3}{[x-2y]^2}$

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

$K_c=\frac{0.243\times (0.729)^3}{(x-2\times 0.243)^2}$

$0.967=\frac{0.243\times (0.729)^3}{(x-2\times 0.243)^2}$

$x=0.80$

concentration of $NH_3$ in the equilibrium mixture = $0.80-2\times 0.243=0.31$

Thus concentration of $NH_3$ in the equilibrium mixture is 0.31 M

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