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2 years ago

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

Chemistry

1 answer:

zysi [14]2 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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The mass of this 25 mL supercritical CO2 sample has a mass of 11.7g

Explanation:

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2 years ago

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1 year ago

In a 0.01 M solution of 1,4-butanedicarboxylic acid, HOOCCH2CH2COOH (Ka1 = 2.9 x 10-5,

azamat

Answer:

The correct answer is B) HOOCCH2CH2COOH(aq)

Explanation:

Both Ka1 and Ka2 are low, so the acid will dissociate only slightly into HOOCCH2CH2COO- ions and even more slightly into -OOCCH2CH2COO- ions. The concentration of hydronium ions (H₃O⁺) will be consequently low. The species that will be in the highest concentration will be HOOCCH2CH2COOH (the weak acid not dissociated).

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2 years ago

what is the concentration of an NaOH solution that requires 50 mL of a 1.25 M H2SO4 solution to neutralize 78.0 ml of NaOH

Dimas [21]

Answer:

The answer is the concentration of an NaOH = 1.6 M

Explanation:

The most common way to solve this kind of problem is to use the formula

C₁ * V₁ = C₂ * V₂

In your problem,

For NaOH

C₁ =?? v₁= 78.0 mL = 0.078 L

For H₂SO₄

C₁ =1.25 M v₁= 50.0 mL = 0.05 L

but you must note that for the reaction of NaOH with H₂SO₄

2 mol of NaOH raect with 1 mol H₂SO₄

So, by applying in above formula

C₁ * V₁ = 2 * C₂ * V₂

(C₁ * 0.078 L) = (2* 1.25 M * 0.05 L)

C₁ = (2* 1.25 M * 0.05 L) / (0.078 L) = 1.6 M

<u>So, the answer is the concentration of an NaOH = 1.6 M</u>

3 0

2 years ago