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Goshia [24]
3 years ago
5

Hi:)) how to do 5? I’m not sure if it’s hydrogen carbonate or carbonate

Chemistry
1 answer:
zysi [14]3 years ago
8 0

Answer:

copper(ll) carbonate

Explanation:

Since the product is a salt which is copper(II) carbonate, water and carbon dioxide, this reaction is an acid + metal carbonate reaction.

Looking at the salt, Cu²⁺ has to be part of the reactants.

Hence, the missing compound there has to be copper(ll) carbonate, CuCO₃.

The balanced chemical equation would be:

CuCO₃ + 2HNO₃➙ Cu(NO₃)₂ +H₂O +CO₂

P.s. You left out CO₂ as a product in Q2 ;)

Just a recap of the main reactions you would've learnt:

1) Acid + base/ alkali ➙ salt + water

2) Acid + metal ➙ salt + hydrogen gas

3) Acid + metal carbonate ➙ salt + H₂O + CO₂

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Answer:

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Explanation:

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5 0
3 years ago
Suppose 0.981 g of iron (II) iodide is dissolved in 150. mL of a 35.0 m M aqueous solution of silver nitrate. Calculate the fina
yaroslaw [1]

Answer:

Final molarity of iodide ion C(I-) = 0.0143M

Explanation:

n = (m(FeI(2)))/(M(FeI(2))

Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol

So n = 0.981/309.85 = 0.0031 mol

V(solution) = 150mL = 0.15L

C(AgNO3) = 35mM = 0.035M = 0.035m/L

n(AgNO3) = C(AgNO3) x V(solution)

= 0.035 x 0.15 = 0.00525 mol

(AgNO3) + FeI(3) = AgI(3) + FeNO3

So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol

C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M

8 0
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