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lisabon 2012 [21]
3 years ago
5

The symbol Na represents a sodium atom that has lost an electron. True False

Chemistry
1 answer:
IgorLugansk [536]3 years ago
3 0
False.
Na does represent sodium, but NOT when it lost an electron. Na+ represents sdoium when it lost its electron.

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Choose the aqueous solution below with the lowest freezing point. These are all solutions of nonvolatile solutes and you should
atroni [7]

Answer:

(NH_4)_3PO_4 0.075 m solution has the lowest freezing point.

Explanation:

Depression in freezing point is given by:

\Delta T_f=T-T_f

\Delta T_f=K_f\times m

\Delta T_f=iK_f\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in Kg}}

where,

\Delta T_f =Depression in freezing point

K_f = Freezing point constant of solvent

1 - van't Hoff factor

m = molality

According question, molality of all the solutions are same and are in prepared with same solvent. So, values of molality and K_f will remain the same and will not effect the freezing point of the solution.

The lowering in freezing point will now depend upon van't Hoff factors of the solutions. Higher the value of van'Hoff factor more will be the lowering in freezing point of the solution.

The van't Hoff factor of KNO_2 solution = i_1=2

The van't Hoff factor of LiCN solution = i_2=2

The van't Hoff factor of (NH_4)_3PO_4 solution = i_3=4

The van't Hoff factor of NaI solution = i_4=2

The van't Hoff factor of NaBrO_3 solution = i_5=2

The solution of ammonium phosphate has the highest values of van't Hoff factor which will result in maximum lowering of the freezing point of the solution.

Hence,(NH_4)_3PO_4 0.075 m solution has the lowest freezing point.

4 0
3 years ago
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How does Barium obey the octet rule when reacting
Kryger [21]

Answer:

it gives up electrons B. because the bond between the electrons is not very strong due to how few of them there are

Explanation:

7 0
3 years ago
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5. The reaction of magnesium oxide with hydrochloric acid carried out in a calorimeter caused the
saul85 [17]

Answer:

kek w

Explanation:

kek w

5 0
3 years ago
What would be the major product obtained from hydroboration–oxidation of the following alkenes?
zmey [24]

Answer:

a. 3-methylbutan-2-ol

b. 2-methylcyclohexan-1-ol

Explanation:

For this reaction, we must remember that the hydroboration is an <u>"anti-Markovnikov" reaction</u>. This means that the "OH" will be added at the <em>least substituted carbon of the double bond.</em>

In the case of <u>2-methyl-2-butene</u>, the double bond is between carbons 2 and 3. Carbon 2 has two bonds with two methyls and carbon 3 is attached to 1 carbon. Therefore <u>the "OH" will be added to carbon three</u> producing <u>3-methylbutan-2-ol</u>.

For 1-methylcyclohexene, the double bond is between carbons 1 and 2. Carbon 1 is attached to two carbons (carbons 6 and 7) and carbon 2 is attached to one carbon (carbon 3). Therefore<u> the "OH" will be added to carbon 2</u> producing <u>2-methylcyclohexan-1-ol</u>.

See figure 1

I hope it helps!

8 0
3 years ago
Please hurry Participants in an informal discussion would most likely
Katyanochek1 [597]
The answer is A. use casual, friendly conversation.
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