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3 years ago

A reaction requires 15.0 g of ammonium chloride. What volumeof a

Chemistry

1 answer:

Lubov Fominskaja [6]3 years ago

7 0

Answer:

103 cm³

Explanation:

A reaction requires 15.0 g of ammonium chloride. The percent by mass of ammonium chloride in the solution is 14%. The mass of the solution required is:

15.0 g ammonium chloride × (100 g solution/ 14 g ammonium chloride) = 107 g solution

The density of the solution is 1.04 g/cm³. The volume of solution corresponding to 107 g is:

107 g × (1 cm³/ 1.04 g) = 103 cm³

In 103 cm³ of a 14% solution, there will be 15.0 g of ammonium chloride.

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What<br>was the initial volume of the hydrogen in cm3?

svetlana [45]

Answer:

255.51cm3

Explanation:

Data obtained from the question include:

V1 (initial volume) =?

T1 (initial temperature) = 50°C = 50 + 273 = 323K

T2 (final temperature) = - 5°C = - 5 + 237 = 268K

V2 (final volume) = 212cm3

Using the Charles' law equation V1/T1 = V2/T2, the initial volume of the gas can be obtained as follow:

V1/T1 = V2/T2

V1/323 = 212/268

Cross multiply to express in linear form

V1 x 268 = 323 x 212

Divide both side by 268

V1 = (323 x 212)/268

V1 = 255.51cm3

Therefore, the initial volume of the gas is 255.51cm3

5 0

3 years ago

Read 2 more answers

Identify the products formed in this Brønsted-Lowry reaction.<br><br> HPO2−4+NO−2↽−−⇀acid+base

klemol [59]

Answer: The products formed in this Bronsted-Lowry reaction are $HNO_{2}$ and $PO^{2-}_{4}$ .

Explanation:

According to Bronsted-Lowry, acids are the species which donate hydrogen ions to another specie in a chemical reaction.

Bases are the species which accept a hydrogen ion upon chemical reaction.

For example, $HPO^{2-}_{4} + NO^{-}_{2} \rightleftharpoons HNO_{2} + PO^{2-}_{4}$

Here, the products formed in this Bronsted-Lowry reaction are $HNO_{2}$ and $PO^{2-}_{4}$ .

Thus, we can conclude that the products formed in this Bronsted-Lowry reaction are $HNO_{2}$ and $PO^{2-}_{4}$ .

5 0

3 years ago

How many milligrams of sodium sulfide are needed to completely react with 25.00 ml of a 0.0100 m aqueous solution of cadmium nit

NARA [144]

Na₂S(aq) + Cd(NO₃)₂(aq) = CdS(s) + 2NaNO₃(aq)

v=25.00 mL
c=0.0100 mmol/mL
M(Na₂S)=78.046 mg/mmol

n(Na₂S)=n{Cd(NO₃)₂}=cv

m(Na₂S)=M(Na₂S)n(Na₂S)=M(Na₂S)cv

m(Na₂S)=78.046*0.0100*25.00≈19.5 mg

5 0

3 years ago

Read 2 more answers

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

Brrunno [24]

Answer:

Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

6 0

2 years ago

Which process is expected to have an increase in entropy?

olganol [36]

Answer: Option (B) is the correct answer.

Explanation:

Degree of randomness of the molecules of a substance is known as entropy. More is the kinetic energy between the molecules of a substance more will be the degree of randomness.

Therefore, when a substance is present in a gaseous state then it has the maximum entropy. In liquid state, molecules are closer to each other so, there is less randomness between them.

On the other hand, in solid state molecules are much more closer to each other as they arr held by strong intermolecular forces of attraction. Therefore, they have very less entropy.

When liquid water is formed from gaseous hydrogen and oxygen molecules then gas is changing into liquid. So, there is decrease in entropy.

When $N_{2}O_{4}$ decomposes then the reaction will be as follows.

$N_{2}O_{4} \rightarrow 2NO_{2}$

Since, 1 mole is producing 2 moles. This means that degree of randomness is increasing as both the molecules are present in gaseous form.

In formation of a precipitate, aqueous solution is changing into solid state. Hence, degree of randomness is decreasing.

Rusting of iron also leads to the formation of solid as it forms $Fe_{2}O_{3}.xH_{2}O$ .

Thus, we can conclude that decomposition of $N_{2}O_{4}$ gas to $NO_{2}$ gas is the process that is expected to have an increase in entropy.

4 0

3 years ago