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3 years ago

If I have a +2 oxidation number do I give up or accept electrons?

Chemistry

1 answer:

ivanzaharov [21]3 years ago

7 0

Answer:

I will give up the electrons

Explanation:

Consider the following rules:

1. An atom loses electrons if its oxidation number is positive.

2. An atom gains electrons if its oxidation number is negative.

3. An atom neither gains nor loses electrons if its atomic number is zero.

As I have an oxidation number $+2$ that is oxidation number is positive, so, I will give up electrons.

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When carbon is oxidized in a small amount of oxygen, the principle product formed is carbon monoxide (co). when the oxidation oc

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3 years ago

Determine the moles of NH3 that can form from 70.0 grams N2.

Darina [25.2K]

Hey there!

5 moles will be produced.

N₂ has a molar mass of 28.014 g/mol.

Convert 70g to mol:

70 ÷ 28.014 = 2.5

In N₂ there are 2 nitrogen atoms. In NH₃ there is 1 nitrogen atom.

So, there will be twice as many moles of NH₃ because every one molecule of N₂ will produce two molecules of NH₃.

2.5 x 2 = 5 moles

Hope this helps!

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3 years ago

Why do group 7 elements have higher melting and boiling points going down?

Genrish500 [490]

Answer:

this is because the van der waals forces or also known as induced dipole-dipole interactions increase because more electrons are present within an atom as you go down

Explanation:

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3 years ago

Calculate the relative formula/atomic mass of each of the following compounds:

zmey [24]

1. HCl (H:1, Cl 35.5) ---> 1+35.5 = 36.5
2. FeS (Fe:56, S:32) ---> 56+32 = 88
3. Cl2 (Cl:35.5) ---> 35.5 x 2 = 71
4. CaC03 (Ca:40, C: 12, O: 16) ---> 40 + 12 + 3(16) = 100
5. Fe0 (Fe: 56, O: 16) ---> 56+16=72

6 0

3 years ago

I need part 1 and 2 please , just separate answers

Vladimir79 [104]

First, we have to remember the molarity formula:

$M=\text{ }\frac{moles\text{ of solute}}{L\text{ solution}}$

Part 1:

In this case, our solute is sodium nitrate (NaNO3), and we have the mass dissolved in water, then we have to convert grams to moles. For that, we need the molecular weight:

$M.W_{NaNO_3}=\text{ 23+14+16*3= 85 g/mol}$

Then, we calculate the moles present in the solution:

$3.976\text{ g NaNO}_3\text{ * }\frac{1\text{ mol}}{85\text{ g}}=\text{ 0.04678 mol NaNO}_3$

Now, we have the necessary data to calculate the molarity (with the solution volume of 200 mL):

$M=\frac{0.04678\text{ mol}}{200\text{ mL*}\frac{1\text{ L}}{1000\text{ mL}}}=\text{ 0.2339 M}$

The molarity of this solution equals 0.2339 M.

Part 2:

In this case, we have the same amount (in moles and mass) of sodium nitrate, but a different volume of solution, then we only have to change it:

$M=\text{ }\frac{0.04678\text{ mol}}{275\text{ mL *}\frac{1\text{ L}}{1000\text{ mL}}}=\text{ 0.1701 M}$

So, the molarity of this solution is 0.1701 M.

5 0

1 year ago

If I have a +2 oxidation number do I give up or accept electrons?​

If I have a +2 oxidation number do I give up or accept electrons?