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AnnyKZ [126]
3 years ago
5

Que tipos deagua oxigenada hay

Chemistry
1 answer:
Murljashka [212]3 years ago
4 0

Answer:

El agua oxigenada se encuentra en cuatro concentraciones diferentes expresadas “en tanto por ciento”: 8%, 30%, 35% y 50%. Sin embargo, hay quien prefiere hablar de “volúmenes” y no de “porcentajes”.

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Ch4 is the lowest boiling point
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Which season is signaled by warmth and new life after a cold period?
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A weak monoprotic acid is titrated with 0.100 MNaOH. It requires 50.0 mL of the NaOH solution to reach the equivalence point. Af
larisa86 [58]

Explanation:

The given reaction is as follows.

        HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)

Hence, number of moles of NaOH are as follows.

        n = 0.05 L \times 0.1 M

           = 0.005 mol

After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.

             n = 0.025 L \times 0.1 M

                = 0.0025 mol

According to ICE table,

         HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)

Initial:     0.005 mol   0.0025 mol              0                  0

Change: -0.0025 mol  -0.0025 mol        +0.0025 mol

Equibm:   0.0025 mol    0                         0.0025 mol

Hence, concentrations of HA and NaA are calculated as follows.

          [HA] = \frac{0.0025 mol}{V}

        [NaA] = \frac{0.0025 mol}{V}

       [A^{-}] = [NaA] = \frac{0.0025 mol}{V}

Now, we will calculate the pK_{a} value as follows.

          pH = pK_{a} + log \frac{A^{-}}{HA}

       pK_{a} = pH - log \frac{[A^{-}]}{[HA]}

                  = 3.42 - log \frac{\frac{0.0025 mol}{V}}{\frac{0.0025}{V}}

                  = 3.42

Thus, we can conclude that pK_{a} of the weak acid is 3.42.

           

8 0
3 years ago
If the density of pure water is 0.9922 g/mL at 40 ºC, calculate its theoretical molarity at that temperature. Report to 4 sig fi
OleMash [197]
Answer is: theoretical molarity of water is 55.1222 mol/L.<span>
d(H</span>₂O) = 0.9922 g/mL.
M(H₂O) = 2 · Ar(H) + Ar(O) · g/mol.
M(H₂O) = 2 + 16 · g/mol = 18 g/mol.
c(H₂O) = d(H₂O) ÷ M(H₂O).
c(H₂O) = 0.9922 g/mL ÷ 18 g/mol.
c(H₂O) = 0.0551 mol/mL.
c(H₂O) = 0.0551 mol/mL · 1000 mL/L = 55.1222 mol/L.
3 0
3 years ago
A helium balloon with an internal pressure of 1atm and a volume of 4.20 L at 18.0°C is released. What volume will the balloon oc
kari74 [83]

Answer:

7.59 L

Explanation:

- Use combined gas law formula and rearrange.

- Change C to K

- Hope that helped! Please let me know if you need further explanation.

3 0
3 years ago
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