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jok3333 [9.3K]
3 years ago
9

Please help I will mark brainy

Chemistry
1 answer:
Artyom0805 [142]3 years ago
3 0

Answer:

the answer is c I took the test

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Find the percent composition of water in barium chloride dihydrate. Round your answer to the
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Answer: its 15 its none of those

Explanation:

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2 years ago
how many moles of sodium hydroxide (NaOH) are required to completely neutralize 2 mol of nitric acid (HNO3)
Brut [27]

2 moles of sodium hydroxide will be needed.

<h3><u>Explanation</u>:</h3>

Sodium hydroxide is a compound which is a base and nitric acid is the acid. The formula of the nitric acid is HNO3 and that of sodium hydroxide is NaOH.

The reaction between them are

NaOH +HNO3 =NaNO3 +H2O.

So here we can see that 1 mole of sodium hydroxide reacts with 1 mole of nitric acid to produce 1 mole of sodium nitrate and 1 mole of water.

So for 2 moles of nitric acid, 2 moles of sodium hydroxide will be required.

6 0
3 years ago
In a titration of a solution of unknown concentration of calcium hydroxide against hydrochloric acid (concentration 0.9mol/dm3),
bearhunter [10]
Hi dear
I hope that will be helpful

6 0
3 years ago
An industrial chemist studying bleaching and sterilizing prepares several hypoclorite buffers.
Vanyuwa [196]

Answer:

pH is 7.60

Explanation:

Let's think the equations:

HClO + H₂O  ⇄  ClO⁻  +  H₃O⁺

As every weak acid, we make an equilbrium

The salt is dissociated in solution

NaClO → Na⁺  +  ClO⁻

                  HClO    +    H₂O    ⇄    ClO⁻    +     H₃O⁺

Initially       0.3m                            0.35m

We have the moles of acid, and the moles of conjugate base.

Reacts          x                                    X                  X

Some amount has reacted, so I obtained (in equilibrium) the moles of base + that amount, and the same amount for H₃O⁺ (ratio is 1:1)

HClO    +    H₂O    ⇄    ClO⁻    +     H₃O⁺

0.3 - x                        0.35 + x           x

Let's make the expression for Ka

Ka = [ClO⁻] . [H₃O⁺]  /  [HClO]

(we don't add water, because it is included in Ka)

2.9x10⁻⁸ = (0.35+x).x  / (0.3-x)

Ka is in order of 10⁻⁸, I can assume that 0.3-x is 0.3  and 0.35 +x =0.3

2.9x10⁻⁸ = (0.35)x  / (0.3)

(2.9x10⁻⁸ .  0.3) /0.35 = x

2.48x10⁻⁸ = x

This is [H₃O⁺]

For pH = - log [H₃O⁺]

pH = 7.60

3 0
3 years ago
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