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3 years ago

You are given 19.2 liters of H3PO4. How many moles do you have?

Chemistry

1 answer:

Pepsi [2]3 years ago

7 0

Answer:

Explanation:

Do a quick conversion: 1 grams H3PO4 = 0.010204583427424 mole using the molecular. How many grams H3PO4 in 1 mol? You can view more details on each measurement unit: molecular weight of H3PO4 or mol This. The formula weight is simply the weight in atomic mass units of all the atoms in a given formula.

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A type of force where objects touch each other

leonid [27]

Answer:

contact force

Explanation:

I have no idea why

3 0

2 years ago

Which statement best describes organs in an organ system? A. Each organ does part of a larger job. B. Each organ can perform onl

nikklg [1K]

Answer:

The correct answer is - A. Each organ does part of a larger job.

Explanation:

An organ in an organ system of an individual organism is the group of similar tissues that collectively perform a common function in the organ system and play their part in a larger job.

A group of organs makes an organ system to perform a particular but large function in the organism for its survival. An example of the organ in an organ system is the heart in the cardiovascular system. The heart is an organ that pumps the blood out of the heart to the various part of the cardiovascular system such as lungs, arteries, and veins so it can take nutrients and oxygen to various parts carried by the blood.

5 0

3 years ago

In chemistry, what is the definition of the term salt? a compound consisting of an anion from an acid and a cation from a base a

inysia [295]

In chemistry<span>, a </span>salt<span> is an ionic compound that can be formed by the neutralization reaction of an acid and a base.So yes it is</span>

7 0

4 years ago

Read 2 more answers

Which of the following pairs lists a substance that can neutralize H2SO4 and the salt that would be produced from the reaction?

Bogdan [553]

The second option only.

LiOH, Li₂SO₄.

<h3>Explanation</h3>

A base neutralizes an acid when the two reacts to produce water and a salt.

Sulfuric acid H₂SO₄ is the acid here. There are more than one classes of bases that can neutralize H₂SO₄. Among the options, there are:

Metal hydroxides

Ca(OH)₂ and
LiOH.

Metal hydroxides react with sulfuric acid to produce water and the sulfate salt of the metal.

$\text{Ca}(\text{OH})_{\bf 2}+\text{H}_2\text{SO}_4 \to \textbf{Ca}\textbf{SO}_{\bf 4} +{\bf 2}\;\text{H}_2\text{O}$ .

The formula for calcium sulfate $\text{CaSO}_4$ in option A is spelled incorrectly. Why? The charge on each calcium $\text{Ca}^{2+}$ is +2. The charge on each sulfate ion ${\text{SO}_4}^{2-}$ is -2. Unlike $\text{Li}^{+}$ ions, it takes only one $\text{Ca}^{2+}$ ion to balance the charge on each ${\text{SO}_4}^{2-}$ ion. As a result, $\text{Ca}^{2+}$ and ${\text{SO}_4}^{2-}$ ions in calcium sulfate exist on a 1:1 ratio.

$2\;\text{LiOH} +\text{H}_2\text{SO}_4 \to \text{Li}_2\text{SO}_4 + 2\;\text{H}_2\text{O}$ .

Ammonia, NH₃

Ammonia NH₃ can also act as a base and neutralize acids. NH₃ exists as NH₄OH in water:

$\text{NH}_3 + \text{H}_2\text{O} \to \textbf{NH}_{\bf 4}\text{OH}$ .

The ion ${\text{NH}_4}^{+}$ acts like a metal cation. Similarly to the metal hydroxides, NH₃ (or NH₄OH) neutralizes H₂SO₄ to produce water and a salt:

$2\;\textbf{NH}_{\bf 4}\text{OH}+ \text{H}_2\text{SO}_4 \to (\textbf{NH}_{\bf 4})_2\text{SO}_4+2\;\text{H}_2\text{O}$ .

The formula of the salt (NH₄)₂SO₄ in the fourth option spelled the ammonium ion incorrectly.

As part of the salt (NH₄)₂SO₄, the ammonium ion NH₄⁺ is one of the products of this reaction and can't neutralize H₂SO₄ any further.

7 0

3 years ago

What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250

Blababa [14]

The buffer solution target has a pH value smaller than that of pKw (i.e., pH < 7.) The solution is therefore acidic. It contains significantly more protons $\text{H}^{+}$ than hydroxide ions $\text{OH}^{-}$ . The equilibrium equation shall thus contain protons rather than a combination of water and hydroxide ions as the reacting species.

Assuming that $x \; \text{L}$ of the 0.307 $\text{mol} \cdot \text{dm}^{-3}$ sodium hydroxide solution was added to the acetic acid. Based on previous reasoning, $x$ is sufficiently small that acetic acid was in excess, and no hydroxide ion has yet been produced in the solution. The solution would thus contain $0.2000 \times 0.425 - 0.307 \; x = 0.085 - 0.307 \; x$ moles of acetic acid and $0.307 \; x$ moles of acetate ions.

Let $\text{HAc}$ denotes an acetic acid molecule and $\text{Ac}^{-}$ denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.

$\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}$

The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of $10^{4.250}\; \text{mol}\cdot \text{dm}^{-3}$ . There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:

$\begin{array}{lcccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\text{C} & - 10^{-4.250} & & +10^{-4.250} & & +10^{-4.250} \\\text{E} & 0.085 - 10^{-4.250} - 0.307 \; x& & 10^{-4.250} & & 10^{-4.250} + 0.307 \; x\end{array}$

By definition:

$\text{K}_{a} = [\text{H}^{+}] \cdot [\text{Ac}^{-}] / [\text{HAc}]\\\phantom{\text{K}_{a}} = 10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x)$

The question states that

$\text{K}_{a} = 1.75 \times 10^{-5}$

such that

$10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x) = 1.75 \times 10^{-5}\\6.16 \times 10^{-5} \; x = 1.48 \times 10^{-6}\\x = 0.0241$

Thus it takes $0.0241 \; \text{L}$ of sodium hydroxide to produce this buffer solution.

6 0

3 years ago