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3 years ago

Solutions that have more h than oh- ions are

Chemistry

2 answers:

erik [133]3 years ago

6 0

Answer:

c. acids.

Explanation:

Acids are substances that have more positive hydrogen ions (H) or protons (cations or anions) than hydroxyl ions (OH) in an aqueous solution; For this reason, they are known as “proton donors”.

In addition, acids react with bases, forming salts and water in a reaction called a "neutralization reaction".

In addition to the prevalence of hydrogen ions in their composition, acidic substances are colorless, have a strong suffocating odor, sour, acidic or bitter taste, pH below 7. low boiling point, conduct electricity in aqueous medium. and react with metals such as iron, magnesium and zinc.

mestny [16]3 years ago

5 0

They are Acids

when acids are in water they dissociate and release H+ ions into the water

while bases release OH- ions

hope that helps

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The reduction in the freezing point of a solution is inversely proportional to a molal concentration

maks197457 [2]

The statement above about "<span>The reduction in the freezing point of a solution is inversely proportional to a molal concentration" is false. It must be directly proportional</span>

7 0

4 years ago

2. Write the formula or name for the following

Tems11 [23]

Answer:

Diphosphorus pentoxide

Carbon dichloride

BCl3

N2H4

Explanation:

These are all covalent compounds. To name covalent compounds, you add prefixes to the beginning of their names depending on what the subscript is of each element. The prefixes are:

1: Mono

2: Di

3: Tri

4: Tetra

5: Penta

6: Hexa

7: Hepta

8: Octa

9: Nona

10: Deca

For example, since the first one is Phopsphorus with a 2 next to it, you add the prefix Di to it.

If the first element in the compound only has one, meaning no number next to it, you do not say mono. This is why we just say "Carbon" for the second one instead of "Monocarbon."

Finally, you always have to end the second element in the compound with "ide." So, "chlorine" becomes "chloride," "oxygen" becomes "oxide," and so on.

7 0

4 years ago

What is the difference between the melting point of molecular and ionic compounds?

irakobra [83]

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6 0

3 years ago

How many protons, neutrons, and electrons are found in a neutral atom of hydrogen-333?

Otrada [13]

Answer:

Explanation:

A hydrogen atom has an atomic number of 1.

So it's proton number will always be 1.

A neutral hydrogen atom does not have any charge.

So it has the same amount of electrons as protons, which turns out to be also 1.

A neutral hydrogen atom has an atomic mass of ≈

1.00794 amu ≈ 1 amu.

To find the amount of neutrons, we take the proton number and subtract it from the mass number.

1 − 1 = 0

So a neutral hydrogen atom has no neutrons.

In summary,

A hydrogen atom has

1 proton

1 electron

0 neutrons

I hope that helps!

4 0

3 years ago

Questions 3-6 refer to rhe solutions below:

Yanka [14]

Under room temperature where $\text{pK}_w = 14$ :

3.) (A), (B), and (E).

4.) (D).

5.) (B).

<h3>Explanation</h3>

What makes a buffer solution? For a solution to be a buffer, it needs to contain large amounts of a weak acid and its conjugate base ion. Alternatively, the solution may contain large amounts of a weak base and its conjugate acid ion.

Not every one of the five solutions is a buffer solution.

Ethanoic acid CH₃COOH (a.k.a. acetic acid) is a weak acid. pKa = 4.756. CH₃COONa is a salt. It dissolves to produce CH₃COO⁻, which is the conjugate base ion of CH₃COOH. The solution in (A) contains equal number of CH₃COOH and CH₃COO⁻, both at 1.0 M.

Refer to the Henderson-Hasselbalch equation for buffers of weak acids.

$\displaystyle \text{pH} = \text{pK}_a + \log{\frac{[\text{Conjugate Ion}]}{[\text{Weak Acid}]}}$ .

$\displaystyle \log{\frac{[\text{Conjugate Ion}]}{[\text{Weak Acid}]}} =\ln{1} = 0$ .

The pH of the solution in (A) will be the same as the pKa of CH₃COOH. pH = 4.746.

Consider the hydrogen halides:

HF: weak acid.
HCl: strong acid.
HBr: strong acid.

The radius of halogen atoms increases down the group, and hydrogen-halogen bond becomes weaker. It becomes easier for water to break those bonds. As a result, the strength of hydrogen halides increases down the group. HF is the only weak acid among the common hydrogen halides.

Mixing HBr and KBr at equal ratio will be similar to mixing HCl and KCl at the same ratio. All HBr in the solution breaks down into H⁺ and Br⁻. The pH of the solution will depend only on the concentration of HBr.

$\displaystyle [\text{H}^{+}] = [\text{HBr}] \\\phantom{[\text{H}^{+}]}= \frac{n}{V} \\\phantom{[\text{H}^{+}]}= \frac{c(\text{HBr})\cdot V(\text{HBr})}{V(\text{HBr})+V(\text{KBr})}\\\phantom{[\text{H}^{+}]}=\frac{0.100\;\text{L}\times 1.0\;\text{mol}\cdot\text{L}^{-1}}{0.100\;\text{L}+0.100\;\text{L}} \\\phantom{[\text{H}^{+}]}= 0.50\;\text{mol}\cdot\text{L}^{-1}$ .

$\text{pH} = -\log{[\text{H}^{+}] = -\log{0.50} \approx {\bf 0.30}$ .

Similarly to HCl and HBr, HI is also a strong acid. Mixing HI and NaOH at equal ratio will produce a solution of NaI, which is similar to NaCl. The final solution will be neutral. pH = 7 if pKw = 14.

NH₃ is a weak base. NH₄Cl dissolves completely to produce NH₄⁺ and Cl⁻. NH₄⁺ is the conjugate acid of NH₃. The final solution will contain an equal number of NH₃ and NH₄⁺. pKb = 4.75 for ammonia NH₃.

Apply the Henderson-Hasselbalch equation for buffers of weak bases:

$\displaystyle \textbf{pOH} = \text{pK}_b + \log{\frac{[\text{Conjugate Ion}]}{[\text{Weak Base}]}}= 4.75 + \log{1} = 4.75$ .

Note that what this equation gives for buffers of weak bases is the pOH of the solution. pH = pKw - pOH. Assume that pKw = 14. pH = 14 - 4.75 = 9.25.

The solution in (E) will contain about 1.0 M of CH₃COOH. The volume of the solution will be 200 mL.

$n(\text{CH}_3\text{COO}^{-}) = n(\text{NaOH}] = c\cdot V = 0.10\;\text{mol}$ .

$\displaystyle [\text{CH}_3\text{COO}^{-}] = \frac{n}{V} = {0.10}{0.10 + 0.10} = 0.50 \;\text{mol}\cdot\text{L}^{-1}$ .

There's nearly no conjugate base of CH₃COOH. As a result, the solution will not be a buffer, and the Henderson-Hasselbalch Equation will not apply. Refer to an ICE table:

$\begin{array}{c|ccccccc}\text{R}&\text{CH}_3\text{COO}^{-} &+&\text{H}_2\text{O}&\rightleftharpoons &\text{CH}_3\text{COOH}&+&\text{OH}^{-}\\\text{I}&0.50\\\text{C}& -x &&&& +x &&+x\\\text{E} &0.50 - x &&&&x&&x\end{array}$

The value of pKa is large. Ka will be small. the value of $x$ will be much smaller than $0.50$ such that $0.50-x \approx 0.50$ .

The pKa of a weak acid is the same as pKw divided by the pKb of its conjugate base.

$\displaystyle \frac{[\text{CH}_3\text{COOH}]\cdot[\text{OH}^{-}]}{[\text{CH}_3\text{COO}^{-}]} = \text{K}_b(\text{CH}_3\text{COO}^{-}) \\\phantom{\displaystyle \frac{[\text{CH}_3\text{COOH}]\cdot[\text{OH}^{-}]}{[\text{CH}_3\text{COO}^{-}]} }= \frac{\text{K}_w}{\text{K}_a(\text{CH}_3\text{COOH})} \\\phantom{\displaystyle \frac{[\text{CH}_3\text{COOH}]\cdot[\text{OH}^{-}]}{[\text{CH}_3\text{COO}^{-}]}} = \frac{10^{-14}}{1.75\times 10^{-5}} = 5.71\times 10^{-10}$ .

$\displaystyle \frac{x^{2}}{0.50} =5.71\times 10^{-10}$ .

$[\text{OH}^{-}] = x \approx 1.69\times 10^{-5}\;\text{mol}\cdot\text{L}^{-1}$ .

$\text{pH} = \text{pK}_w + \log{[\text{OH}^{-}]} = 9.23$ .

7 0

4 years ago