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2 years ago

3. Do distilled water conduct electricity? What will happen if we add sugar to it and

Chemistry

2 answers:

nikitadnepr [17]2 years ago

8 0

Answer:

It cannot conduct electricity, however adding salt or sugar will make the water have impurities/other substance making it easier to conduct electricity

Explanation:

Distilled water by itself does not contain impurities, thus, it cannot conduct electricity.

When you put salt in water, the water molecules pull the sodium and chlorine ions apart so they are floating freely, increasing the conductivity.

For more information, please refer to the internet :D

Have fun studying, and goodluck!

If you are satisfied with this answer, please rate it or give brainliest.

Elden [556K]2 years ago

8 0

Answer:

• Distillate does not conduct electricity. This is because it lacks ions which may allow flow of electricity.

• If sugar is added to the distillate, it becomes a solution and contains ions which become mobile when electricity is passed through. Due to this mobility, the electrons are moved by these ions hence flow of current.

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Which factor determine the reactivity of alkyl halide

Pavel [41]

Answer:

This reactivity order reflects both the strength of the C–X bond, and the stability of X(–) as a leaving group, and leads to the general conclusion that alkyl iodides are the most reactive members of this functional class.

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3 years ago

If the compound has a molar mass of 156 g/mol, what is its molecular formula?

oksano4ka [1.4K]

I believe that the choices for this question are:
C2H4O2, C4H8O4 CH2O, C6H12O6 C3H6O3, C6H12O6 C2H4O2, C6H12O6

The answer to this based on the molar masses given is:
C2H4O2, C6H12O6

To prove calculate the molar mass:
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C6H12O6 = 6*12 + 12*1 + 6*16 = 180

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3 years ago

What is the volume that you should dilute 30mL of a 12.0M H2SO4 solution to obtain a 0.160M H2SO4 solution?

shutvik [7]

To solve this problem, the dilution equation (M1 x V1 = M2 X V2) must be used. The given values in the problem are M1= 12.0 M, V1= 30 mL, and M2= 0.160 M. To solve for V2,
V2=M1xV1/M2
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3 years ago

If you mix a 25.0mL sample of a 1.20m potassium sulfide solution with 15.0 mL of a 0.900m basium nitrate solution and the reacti

ddd [48]

Answer:

Limiting reagent: barium nitrate

Theoretical yield: 2.29 g BaS

Percent yield: 87%

Explanation:

The corrected balanced reaction equation is:

K₂S + Ba(NO₃)₂ ⇒ 2KNO₃ + BaS

The amount of potassium sulfide added is:

(25.0 mL)(1.20mol/L) = 30 mmol

The amount of barium nitrate added is:

(15.0mL)(0.900mol/L = 13.5 mmol

Since the reactant react in a 1:1 molar ratio, barium nitrate is the limiting reagent. If all of the limiting reagent reacts, the amount of barium sulfide produced is:

(13.5 mmol Ba(NO₃)₂)(BaS/Ba(NO₃)₂ ) = 13.5 mmol BaS

Converting this amount to grams gives the theoretical yield of BaS (molar mass 169.39 g/mol).

(13.5 mmol)(169.39 g/mol)(1g/1000mg) = 2.29 g BaS

The percent yield is calculated as follows:

(actual yield) / (theoretical yield) x 100%

(2.0 g) / (2.29 g) x 100% = 87%

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