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Sauron [17]
2 years ago
8

A substance with a high [H ] would likely have which additional characteristics? a high [OHâ€"" ] and a high pOH a low [OHââ

‚¬"" ] and a high pOH a high [OHâ€"" ] and a low pOH a low [OHâ€"" ] and a low pOH.
Chemistry
1 answer:
viktelen [127]2 years ago
4 0

The negative log function that determines the acidity or alkalinity by hydronium ion concentration is called pH.

The substance having high \rm [H] will have:

Option B. A low \rm [OH ^{-}] and a high \rm pOH

This characteristic can be explained as:

  • The concentrations of \rm [H^{+}] and \rm [OH ^{-}] are inversely dependent on each other so when the concentration of  raises then the concentration of \rm [OH ^{-}] drops and vice versa.

  • The pH of a solution or substance is calculated with the help of:

\rm  pH  = \rm - log  \rm [H^{+}]

From the formula, it can be deduced that when the concentration of \rm [H^{+}] is high then the pH has a low value and it means that the solution is acidic.

  • pH can also be written as:

\rm pH + pOH &= 14 \\\\\rm pOH &= 14 - pH

From this formula we can that when the value of pH is less then the value of pOH will be increased and vice versa.

Therefore, when pOH is in high concentration then \rm [OH ^{-}] is low.

To learn more about pH and pOH follow the link:

brainly.com/question/13557815

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7 0
3 years ago
The flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.160 M NaOH. It requires 18.
olchik [2.2K]

Answer:

Approximately 0.291\; \rm M (rounded to two significant figures.)

Explanation:

The unit of concentration \rm M is the same as \rm mol \cdot L^{-1} (moles per liter.) On the other hand, the volume of both the \rm NaOH solution and the original \rm HCl solution here are in milliliters. Convert these two volumes to liters:

  • V(\mathrm{NaOH}) = 18.2\; \rm mL = 18.2 \times 10^{-3}\; \rm L = 0.0182\; \rm L.
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Calculate the number of moles of \rm NaOH in that 0.0182\; \rm L of 0.160\; \rm M solution:

\begin{aligned} n(\mathrm{NaOH}) &= c(\mathrm{NaOH})\cdot V(\mathrm{NaOH})\\ &= 0.160\; \rm mol \cdot L^{-1} \times 0.0182\; \rm L \approx 0.00291\; \rm mol\end{aligned}.

\rm HCl reacts with \rm NaOH at a one-to-one ratio:

\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l).

Coefficient ratio:

\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1.

In other words, one mole of \rm NaOH would neutralize exactly one mole of \rm HCl. In this titration, 0.291\; \rm mol of \rm NaOH\! was required. Therefore, the same amount of \rm HC should be present in the original solution:

\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}.

Calculate the concentration of the original \rm HCl solution:

\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M.

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Two elements are studied. One has atomic number X and one has atomic number X+2. It is known that element X is a halogen. To wha
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Answer:

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If element X is a halogen, then it belongs to the group 17 (or VIIA, under a different notation).

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