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fgiga [73]
3 years ago
5

An ion with a -2 charge has? A) one missing protein B) two missing protons c) two missing electrons D) two extra electrons

Chemistry
2 answers:
Law Incorporation [45]3 years ago
8 0
The correct answer is d
Andreyy893 years ago
3 0

Answer:

D.

Explanation:

It would not be gain because it would say +6 instead of -2. Having a -2 charge means it wants to loose 2 electrons so it can form a full shell of 8 electrons. It wants to be inert (stable).

- Hope that helps! Please let me know if you need further explanation.

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A 35.0 ml sample of 0.225 m hbr was titrated with 42.3 ml of koh. What is the concentration of the koh?
Lemur [1.5K]

Answer:

The concentration of KOH is 0.186 M

Explanation:

First things first, we need too write out the balanced equation between HBr and KOH.

This is given as;

KOH (aq) + HBr (aq) → KBr (aq) + H2O (l)

From the reaction above, we can tell that it takes 1 mole of KOH to react with 1 mole of HBr.

We use the acid base formular in calculating unknown concentrations. This is given as;

\frac{CaVa}{CbVb}  = \frac{na}{nb}

where;

Ca = Concentration of acid

Va = Volume of acid

Cb = Concentration of base

Vb = Volume of base

na = Number of moles of acid

nb = Number of moles of base

KOH is the base and HBr is acid.

Hence;

Ca = 0.225

Va = 35

Cb = ?

Vb = 42.3

na = 1

nb = 1

Making Cb subject of formular we have;

Cb = \frac{CaVaNb}{VbNa}

Cb = (0.225 * 35 * 1) / (42.3 * 1)

Cb = 0.186 M

5 0
3 years ago
An object was measured by a worker as 35.6 cm long, however, the manufacturer specifications list the length of the object at 35
Anon25 [30]
The answer for this question is 1.71%
7 0
3 years ago
Which salt is produced when copper oxide reacts with hydrochloric acid?
Brilliant_brown [7]
It forms copper chloride
5 0
3 years ago
Read 2 more answers
At 25 °C, what is the hydroxide ion concentration, [OH−] , in an aqueous solution with a hydrogen ion concentration of [H+]=4.8×
algol [13]

[OH⁻]= 1 x 10⁻¹⁴ : 4.8 x 10⁻⁴ = 2.083 x 10⁻¹¹

3 0
1 year ago
What volume would be needed to prepare 375 mL of a .45 M CaCl2 using only a solution of 1.0 M CaCl2 and water?
Neporo4naja [7]

Answer:

168.75 ml

Explanation:

M1V1=M2V2

375ml*.45M=1M*V2

8 0
3 years ago
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