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mezya [45]
3 years ago
10

calculate the heat required in joules to convert 18.0 grams of water ice at a temperature of -20 c to liquid water at the normal

boiling point of water given specific heat of ice 2.09
Chemistry
1 answer:
KIM [24]3 years ago
7 0

Answer:

Explanation:

Heat required to convert ice to ice at 0⁰C

= mass x specific heat x rise in temperature

= 18 x 2.09 x 20

= 752.4 J .

heat required to convert ice at 0⁰C to water at 0⁰C

mass x latent heat of fusion

= 18 x 336

= 6048 J

Heat required to increase the temperature of water to 100⁰C

= 18 x 4.2 x 100

= 7560 J

Total heat required

7560 + 6048 + 752.4

= 14360.4 J

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Write a net ionic equation to show how codeine, c18h21o3n, behaves as a base in water.
Lera25 [3.4K]

Actually, the ionic equation for this is a reversible equation since codeine is a weak base. Any weak base or weak acids do not completely dissociate which makes them a reversible process. The ionic equation for this case is:

<span>C18H21O3N  +  H3O+  </span><=> C18H21O3NH+ +  H2O 

7 0
3 years ago
You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30
dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula:\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{1}{3})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = 0.012345

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

3 0
3 years ago
I performed an experiment and mixed copper nitrate and potassium iodide. When they reacted, they formed a precipitate, even thou
Alexus [3.1K]

Answer:

2Cu2^+ + 2I^- ----> 2Cu^+ + I2

Explanation:

The reaction performed in the experiment is;

2 Cu(NO3)2 + 4 KI → 2 CuI (s) + 4 KNO3 + I2

The iodide ions reduces Cu^2+ to Cu^+ which is insoluble in water hence the precipitate. This is so because iodine is a good oxidizing agent seeing that it requires one electron to fill its outermost shell. Potassium on the other hand is a good reducing agent since it easily looses its one electron.

The oxidation - reduction equation is as follows;

2Cu2^+ + 2e ----> 2Cu^+ reduction half equation

2I^- ----> I2 + 2e. Oxidation half equation

Balanced redox reaction equation;

2Cu2^+ + 2I^- ----> 2Cu^+ + I2

3 0
3 years ago
How many minutes are in 17 years?
Anarel [89]

Explanation:

8935200 mins plzzzźzzz mark it brainest

3 0
3 years ago
Chemical reactions forming ions do not balance electrically. <br><br> True or False
FromTheMoon [43]

Answer: true

Explanation:

4 0
3 years ago
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