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2 years ago

Give an example of a substance changing temperature and what happens to the motion of molecules as the temperature changes?

Chemistry

1 answer:

Svetradugi [14.3K]2 years ago

6 0

Answer:

When water freezes, the molecules slow down until the water becomes a solid and the molecules stop moving.

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Show the difference between a saturated hydrocarbon and an unsaturated hydrocarbon.

harkovskaia [24]

Saturated hydrocarbons consists of C-C single bond whereas Unsaturated hydrocarbons consists C-C double/triple bond.

4 0

2 years ago

Sodium sulfate is slowly added to a solution containing 0.0500 M Ca 2 + ( aq ) and 0.0390 M Ag + ( aq ) . What will be the conce

levacccp [35]

The given question is incomplete. The complete question is as follows.

Sodium sulfate is slowly added to a solution containing 0.0500 M $Ca^{2+}(aq)$ and 0.0390 M $Ag^{+}(aq)$ . What will be the concentration of $Ca^{2+}$ (aq) when $Ag_{2}SO_{4}(s)$ begins to precipitate? What percentage of the $Ca^{2+}(aq)$ can be separated from the Ag(aq) by selective precipitation?

Explanation:

The given reaction is as follows.

$Ag_{2}SO_{4} \rightleftharpoons 2Ag^{+} + SO^{2-}_{4}$

$[Ag^{+}]$ = 0.0390 M

When $Ag_{2}SO_{4}$ precipitates then expression for $K_{sp}$ will be as follows.

$K_{sp} = [Ag^{+}]^{2}[SO^{2-}_{4}]$

$1.20 \times 10^{-5} = (0.0390)^{2} \times [SO^{2-}_{4}]$

$[SO^{2-}_{4}]$ = 0.00788 M

Now, equation for dissociation of calcium sulfate is as follows.

$CaSO_{4} \rightleftharpoons Ca^{2+} + SO^{2-}_{4}$

$K_{sp} = [Ca^{2+}][SO^{2-}_{4}]$

$4.93 \times 10^{-5} = [Ca^{2+}] \times 0.00788$

$[Ca^{2+}]$ = 0.00625 M

Now, we will calculate the percentage of $Ca^{2+}$ remaining in the solution as follows.

$\frac{0.00625}{0.05} \times 100$

= 12.5%

And, the percentage of $Ca^{2+}$ that can be separated is as follows.

100 - 12.5

= 87.5%

Thus, we can conclude that 87.5% will be the concentration of $Ca^{2+}(aq)$ when $Ag_{2}SO_{4}(s)$ begins to precipitate.

4 0

2 years ago

If 250.0 g of water at 30.0 °C cool to 5.0 °C, how many kilojoules of energy did the water lose?

Y_Kistochka [10]

Answer:

-26.125 kj

Explanation:

Given data:

Mass of water = 250.0 g

Initial temperature = 30.0°C

Final temperature = 5.0°C

Amount of energy lost = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

ΔT = 5.0°C - 30.0°C

ΔT = -25°C

Specific heat of water is 4.18 j/g.°C

Now we will put the values in formula.

Q = m.c. ΔT

Q = 250.0 g × 4.18 j/g.°C × -25°C

Q = -26125 j

J to kJ

-26125 j ×1 kj /1000 j

-26.125 kj

5 0

2 years ago

Which isomer of 1-bromo-4-tert-butylcyclohexane reacts faster when refluxed with potassium tert-butoxide?

grigory [225]

Answer:

See explanation.

Explanation:

Since potassium tert-butoxide is a strong bulky base, we expect that an elimination reaction predominates.

This reaction is expected to occur by E2 mechanism. The cis isomer of 1-bromo-4-tert-butylcyclohexane is known to react faster by E2 mechanism because it fulfills the anticoplanar arrangement required for for E2 mechanism, where the leaving group and adjacent proton must be anti to each other and in the same plane. The trans isomer can not fulfill this requirement.

7 0

2 years ago

7. What is the molarity, M, of the dilution solution of KBr, given:

sineoko [7]

Answer:

Hello There!!

Explanation:

I believe the answer is c. Volume of the dilute solution is 100.00 mL.

hope this helps,have a great day!!

~Pinky~

7 0

2 years ago

Read 2 more answers