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Inga [223]
3 years ago
13

The monoprotic acid from among the following is

Chemistry
2 answers:
oksian1 [2.3K]3 years ago
3 0

Monoprotic acid are acids having only one hydrogen atoms after dissociation into ions from its compound. The monoprotic acid from among the following is HCl. The answer is letter D. HCl → H+ + Cl-. Note that there is only one H+ ion upon dissociation.

Marianna [84]3 years ago
3 0

Answer: Option (d) is the correct answer.

Explanation:

An acid which gives only one hydrogen ion or H^{+} ions upon dissociation in an aqueous solution is known as a monoprotic acid.

Whereas when an acid gives two hydrogen or H^{+} ions upon dissociation in an aqueous solution is known as a diprotoc acid.

And when an acid gives three hydrogen or H^{+} ions upon dissociation in an aqueous solution is known as a triprotoc acid.

Therefore dissociation of the given acids in an aqueous solution will be as follows.

  H_{2}CO_{3} \rightarrow 2H^{+} + CO^{2-}_{3}

  H_{2}SO_{4} \rightarrow 2H^{+} + SO^{2-}_{4}

  H_{3}PO_{4} \rightarrow 3H^{+} + PO^{2-}_{4}

  HCl \rightarrow H^{+} + Cl^{-}

Hence, we can conclude that out of the given options, HCl is the monoprotic acid.

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A molecule is polar when:
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The answer would be B. One region of the molecule has a small negative charge while another region has a small positive charge. However usually in polar bonds, charges or bond between the atoms are unequal (as opposed to having small equal charges).
6 0
3 years ago
For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coeffici
Brut [27]

<u>Question:</u>

For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coefficient for 0.124 b HCl solution if E=0.342 V at 298 K

<u>Answer:</u>

The mean activity coefficient for HCl solution is 0.78.

<u>Explanation:</u>

Activity coefficient is defined as the ratio of any chemical activity of any substance with its molar concentration. So in an electrochemical cell, we can write activity coefficient as γ. The equation for determining the mean activity coefficient is

              E=E_{0}-0.0514 \mathrm{V} \ln \gamma

As we know that E_{0} = 0.22 V and E = 0.342 V, the equation will become

             0.342 V+0.0514 V \ln (0.124)=0.22 V-0.0514 V \ln \gamma

             0.342 V-0.222 V=-0.0514 V(\ln \gamma+\ln 0.124)

             0.12 \mathrm{V}=-0.0514 \mathrm{V}(\ln \gamma+\ln 0.124)

             \frac{0.12}{0.0514}=-\ln (0.124 \gamma)

             -2.3346=\ln (0.124 \gamma)

             e^{-2.3346}=0.124 \gamma

             0.0968=0.124 \gamma

             \gamma=\frac{0.0968}{0.124}=0.78

So, the mean activity coefficient is 0.78.

6 0
3 years ago
A gas sample occupies 200 mL at 760 mm Hg. What volume does the gas occupy at 400 mm Hg?
astra-53 [7]

Answer:

201.9

Explanation:

when you divided 760 with 400 yo get 19.0 the add it with 200 you get that answer

3 0
3 years ago
What two things do all matter have in common?
ELEN [110]

Answer:

mass and occupies space

Explanation:

matter is made up of mass and mass takes up space

3 0
3 years ago
Read 2 more answers
4. Magnesium and oxygen undergo a chemical reaction to
erastovalidia [21]

Answer:

About 16.1 grams of oxygen gas.

Explanation:

The reaction between magnesium and oxygen can be described by the equation:
\displaystyle 2\text{Mg} + \text{O$_2$} \longrightarrow 2\text{MgO}

24.4 grams of Mg reacted with O₂ to produce 40.5 grams of MgO. We want to determine the mass of O₂ in the chemical change.

Compute using stoichiometry. From the equation, we know that two moles of MgO is produced from every one mole of O₂. Therefore, we can:

  1. Convert grams of MgO to moles of MgO.
  2. Moles of MgO to moles of O₂
  3. And moles of O₂ to grams of O₂.

The molecular weights of MgO and O₂ are 40.31 g/mol and 32.00 g/mol, respectively.

Dimensional analysis:

\displaystyle 40.5\text{ g MgO} \cdot \frac{1\text{ mol MgO}}{40.31\text{ g MgO}} \cdot \frac{1\text{ mol O$_2$}}{2\text{ mol MgO}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}

In conclusion, about 16.1 grams of oxygen gas was reacted.

You will obtain the same result if you compute with the 24.4 grams of Mg instead:

\displaystyle 24.4\text{ g Mg}\cdot \frac{1\text{ mol Mg}}{24.31\text{ g Mg}} \cdot \frac{1\text{ mol O$_2$}}{1\text{ mol Mg}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}

3 0
2 years ago
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