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4 years ago

The periodic chart says that f can take on a charge of -1 to become f-1, how does this happen?

1 answer:

7 0

These electrons can either be lost to another atom, or the valence shell may have room ... want to gain or lose an electron, it will never happen with an isolated atom. ... that make it very easy to determine the charge directly from the periodic table. ... He0. Li+. Be+2. B+3. C-4. N-3. O-2. F-. Ne0. Na+. Mg+2. Al+3. Si-4. P-3. S-2.

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Rank the organic compounds from most soluble to least soluble. To rank items as equivalent, overlap them.

denpristay [2]

The answer to your questions is as follows:

most soluble
>CH3CH2OH
>CH3OCH3
>CH3CH3
least soluble

I hope my answer has come to your help. God bless and have a nice day ahead!

7 0

3 years ago

What the answer I’m on a midterm and can’t seem to find it help

Sonja [21]

Answer:

Sorry I’m not rlly sure but maybe the 2nd or the last

Explanation:

8 0

3 years ago

- Equilibrium shifts for slightly soluble compounds The reaction for the formation of a saturated solution of silver bromide (Ag

RideAnS [48]

Answer:

1) Favor formation of reactant

2) Favor formation of product

3) Favor formation of product

4) Favor formation of product

5) Favor formation of product

Explanation:

The reaction is $AgBr(s)--->Ag^{+}(aq)+Br^{-}(aq)$

The reaction is endothermic

1. Adding hydrobromic acid (HBr)

It will increase the amount of bromide ion. Bromide ion is one of the product so it will favor formation of reactant.

2. Lowering the concentration of bromide (Br-) ions

As we are decreasing the the concentration of product it will favor formation of more products

3. Heating the concentrations

If we are heating the mixture, it will increase the temperature and it will favor endothermic reaction. Thus will favor formation of product.

4. Adding solid silver bromide (AgBr)

If we are adding silver bromide it means we are increasing the concentration of reactant, it will favor formation of products

5. Removing silver (Ag+) ions

Removing silver ions means we are decreasing the concentration of product thus it will favor formation of products.

3 0

3 years ago

What type of radioactive decay will the isotopes 13B and 188Au most likely undergo?

scoundrel [369]

Answer:

b. Beta emission, beta emission

Explanation:

A factor to consider when deciding whether a particular nuclide will undergo this or that type of radioactive decay is to consider its neutron:proton ratio (N/P).

Now let us look at the N/P ratio of each atom;

For B-13, there are 8 neutrons and five protons N/P ratio = 8/5 = 1.6

For Au-188 there are 109 neutrons and 79 protons N/P ratio = 109/79=1.4

For B-13, the N/P ratio lies beyond the belt of stability hence it undergoes beta emission to decrease its N/P ratio.

For Au-188, its N/P ratio also lies above the belt of stability which is 1:1 hence it also undergoes beta emission in order to attain a lower N/P ratio.

8 0

3 years ago

Sn + 2 H2SO4 → SnSO4 + SO2 + 2 H2O If 219.65 grams of SnSO4 are produced, how many moles of H2SO4 were reacted?

anygoal [31]

Answer:

2.05moles

Explanation:

The balanced chemical equation in this question is as follows;

Sn + 2H2SO4 → SnSO4 + SO2 + 2H2O

Based on the above equation, 2 moles of H2SO4 reacted to produce 1 mole of SnSO4

However, the mass of SnSO4 produced is 219.65 grams. Using mole = mass/molar mass, we can find the number of moles of SnSO4 produced.

Molar mass of SnSO4 where Sn = 118.7, S = 32, O = 16

= 118.7 + 32 + 16(4)

= 150.7 + 64

= 214.7g/mol

mole = 219.65/214.7

mole = 1.023mol

Therefore, if 2 moles of H2SO4 reacted to produce 1 mole of SnSO4

1.023 mol of SnSO4 produced will cause: 1.023 × 2/1

= 2.046moles of H2SO4 to react.

8 0

3 years ago