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2 years ago

PLEASE HELP! NO LINKS! CHEMISTRY!

Chemistry

2 answers:

Bingel [31]2 years ago

6 0

Answer:

Solution A is 1,000 times more acidic than Sol. B

Explanation:

for pH values we use scientific notation:

-log10 c (where c is the hydrogen ion concentration) is used to notate pH value (think of it as a unit)

ie:

10^-2 is sol A 10^-5 is sol B

5-2 is 3

10^-3 = 1000

there's a diff of 1,000 between the solutions.

aleksley [76]2 years ago

5 0

Answer: solution A is more acidic as it closer to 0 on the pH scale.It’s also three times more acidic that solution B as when the [H+] concentration for both solutions is calculated it’s found that solution A is three times more concentrated than solution B.

Explanation:

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A sealed balloon occupies a volume of 250 mL at an unknown temperature. When the temperature is altered to 30 C, the balloon's v

Allushta [10]

Answer:

Initial temperature, T1 = 15°C

Explanation:

Given the following data;

Initial volume, V1 = 250 mL to Liters = 250/1000 = 0.25 L

Final temperature, T2 = 30°C

Final volume, V2 = 0.5 L

To find the initial temperature T1, we would use Charles' law.

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles is given by;

$\frac {V}{T} = K$

$\frac{V1}{T1} = \frac{V2}{T2}$

Making T1 the subject of formula, we have;

$T_{1}= \frac{V_{1}}{V_{2}} * T_{2}$

Substituting into the formula, we have;

$T_{1}= \frac{0.25}{0.5} * 30$

$T_{1}= 0.5 * 30$

<em>Initial temperature, T1 = 15°C</em>

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2 years ago

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2 years ago

Calculate the mass percent of HOCH 2CH 2OH in a solution made by dissolving 3.2 g of HOCH 2CH 2OH in 43.5g of water.

grin007 [14]

Answer:

$\%m/m=6.85\%$

Explanation:

Hello,

In this case, we are asked to compute the by mass percent for the given 3.2 g of ethylene glycol in 43.5 g of water. In such a way, since the by mass percent is computed as follows:

$\%m/m=\frac{m_{solute}}{m_{solute}+m_{solvent}} *100\%$

Whereas the solute is the ethylene glycol and the solvent the water, therefore we obtain:

$\%m/m=\frac{3.2g}{3.2g+43.5g} *100\%\\\\\%m/m=6.85\%$

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