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IRINA_888 [86]
3 years ago
14

What is wrong with the name (given in parentheses or brackets) for each of the following compounds: (a) BaCl2(barium dichloride)

, (b) Fe2O3 liron(ll) oxide), (c) CsNO2 (cesium nitrate), (d) Mg(HCO3)2 (magnesium(ll) bicarbonate)?
Chemistry
1 answer:
Aleksandr-060686 [28]3 years ago
3 0

Answer:

a) barium chloride

b) iron (III) oxide

c) cesium nitrite

d) magnesium bicarbonate

Explanation:

Inorganic compounds are named based on the following conventions:

Metal cation + Non-metal anion

The name of the metal is retained as such, whereas the anion name is followed by the suffix 'ide'

For polyatomic anions the corresponding name is added to the name of the metal

Only for transition metals, the specific oxidation state exhibited by the metal cation is mentioned in roman numerals.

a) BaCl2

Ba, Barium cation +  chloride anion

Name = Barium chloride

b) Fe2O3

Fe is a transition metal. In this compound the oxidation state of Fe is +3 and not +2. Hence the name is iron (III) oxide

c) CsNO2

Cs, cesium cation + NO2-, nitrite anion

Name = cesium nitrite

d) Mg(HCO3)2

Mg, magnesium cation + HCO3-, bicarbonate anion

Name = magnesium bicarbonate

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At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react
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At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

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We first need the reaction.

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N_{2 (g)} + O_{2 (g)} ⇄ 2NO_{(g)}

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no N_{2 (g)} nor  O_{2 (g)} present. Immediately, N_{2 (g)} andO_{2 (g)} are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [N_{2 (g)}]=0   ;     [O_{2 (g)} ]= 0    ; [NO_{(g)}]=0.60M

C: [N_{2 (g)}]=+x   ;     [O_{2 (g)} ]= +x    ; [NO_{(g)}]=-2x

E: [N_{2 (g)}]=0+x   ;     [O_{2 (g)} ]= 0+x   ; [NO_{(g)}]=0.60-2x

Now we can use the constant information:

K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }

4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}

4.10* 10^{-4}= \frac{(0.60-2x)^{2}}{x^{2} }

4.10* 10^{-4} * x^{2}= (0.60-2x)^{2}}

\sqrt{4.10* 10^{-4} * x^{2}}= \sqrt{(0.60-2x)^{2}}}

0.0202 x =0.60 - 2x

2x+0.0202x=0.60

x=\frac{0.60}{2.0202}= 0.30

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

3 0
2 years ago
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