PLZ HELP WILL BRAINLIEST

A solution at 25 degrees Celslus 1.0*10^ -5 MH 3 O^ + . What the concentration of OH^ - in this solution?

dexar [7]

Answer:

D.

Explanation:

-log(1.0x10^-5) = pH

pH + pOH = 14 (rearrange it)

OH- = 10^-pOH = 1.0 x 10^-9

- Hope that helped! Let me know if you need further explantion.

7 0

3 years ago

Which element above has 6 protons in its nucleus?

Vinil7 [7]

B. carbon because it's atomic number is 6

5 0

2 years ago

Read 2 more answers

Given a fixed amount of gas help at a constant pressure, calculate the temperature to which the gas would have to be changed if

PIT_PIT [208]

Answer:

592 K or 319° C

Explanation:

From the statement of Charles law we know that the volume of a given mass of gas is directly proportional to its absolute temperature at constant pressure. Thus;

V1/T1= V2/T2

Initial volume V1 = 1.75 L

Initial temperature T1= 23.0 +273 = 296 K

Final volume V2= 3.50 L

Final temperature T2 = the unknown

T2= V2T1/V1= 3.50 × 296 / 1.75

T2 = 592 K or 319° C

4 0

2 years ago

Julia did an experiment to study the solubility of two substances. She poured 100 mL of water at 20 °C into each of two beakers

Aleks [24]

<u>Answer:</u> The solubility of B is high than the solubility of A.

<u>Explanation:</u>

The solubility is defined as the amount of substance dissolved in a given amount of solvent. More the solute gets dissolved, high will be the solubility and less the solute dissolved, low will be the solubility.

For the given observations:

Mass of undissolved substance of substance A is more than Substance B at every temperature. This implies that less amount of solute gets dissolved in the given amount of solvent.

Hence, substance B has high solubility than substance A.

7 0

3 years ago

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25.0cm3 of s saturated potassium hydroxide is neutralized by 35.0cm3 of hydrogen chloride acid of concentration 0.75 mol/dm3. Ca

Kamila [148]

Answer:

Concentration of the original $\rm KOH$ solution: approximately $1.05\; \rm mol \cdot dm^{-3}$ .

Explanation:

Notice that the concentration of the $\rm HCl$ solution is in the unit $\rm mol\cdot dm^{-3}$ . However, the unit of the two volumes is $\rm cm^{3}$ . Convert the unit of the two volumes to $\rm dm^{3}$ to match the unit of concentration.

$\begin{aligned} V(\mathrm{NaOH}) &= 25.0\; \rm cm^{3} \\ &= 25.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0250\; \rm dm^{3} \end{aligned}$ .

$\begin{aligned} V(\mathrm{HCl}) &= 35.0\; \rm cm^{3} \\ &= 35.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0350\; \rm dm^{3} \end{aligned}$ .

Calculate the number of moles of $\rm HCl$ molecules in that $0.0350\; \rm dm^{3}$ of $0.75\; \rm mol \cdot dm^{3}$ $\rm HCl\!$ solution:

$\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}$ .

$\rm HCl$ is a monoprotic acid. In other words, each $\rm HCl\!$ would release up to one proton $\rm H^{+}$ .

On the other hand, $\rm KOH$ is a monoprotic base. Each $\rm KOH\!$ formula unit would react with up to one $\rm H^{+}$ .

Hence, $\rm HCl$ molecules and $\rm KOH\!$ formula units would react at a one-to-one ratio.

${\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l)$ .

Therefore, that $0.02625\; \rm mol$ of $\rm HCl$ molecules would neutralize exactly the same number of $\rm NaOH$ formula units. That is: $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ .

Calculate the concentration of a $\rm NaOH$ solution where $V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3}$ and $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ :

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}$ .

7 0

2 years ago