Chemistry High School

Which of the following equations violates the law of conservation of mass?

A) 2Al + 6HCl yields 2AlCl3 + 3H2
B) Na2C2O4 + Pb(NO3)2 yields PbC2O4 + 2NaNO3
C) AgNO3 + LiCl yields AgCl + LiNO3
D) 3Ca + 2Cr(NO3)3 yields 3Ca(NO3)3 + 2Cr

Answers

Answer 1
D is the answer.
On the left, there are 3 atoms of Ca, 2 atoms of Cr, 6 atoms of N, and 18 atoms of O.
On the right, there are 3 atoms of Ca, 9 atoms of N, 27 atoms of O, and 2 atoms of Cr.
Answer 2

Answer: Option (D) is the correct answer.

Explanation:

Law of conservation of mass states that energy can neither be created nor it can be destroyed. It can only be transformed from one form to another.

Therefore, in a chemical reaction an equation can only be balanced if the number of reactants equal the number of products because energy required for reactants is utilized in the formation of the products.

For example, [tex]2Al + 6HCl \rightarrow 2AlCl_{3} + 3H_{2}[/tex] is a balanced equation because the number of reactant atoms equal the number of product atoms.

Thus, we can conclude that out of the given options, [tex]3Ca + 2Cr(NO3)_3 \rightarrow 3Ca(NO3)_{3} + 2Cr[/tex] equations violates the law of conservation of mass because number of reactant atoms does not equal the number of product atoms.

Related Questions

Friction and motion occur at the same time.

True
False

Answers

true, friction is the force that opposes motion.

The ph of a 0.55 m aqueous solution ammonia, nh3, at 25.0°c is 11.50. what is the value of kb for nh3?

Answers

Kb is called the equilibrium constant of basicity. These are actually constants for weak acids. The value for ammonia, NH₃, is actually 1.8×10⁻⁵. This is used to know how much moles of ammonia could react in an equilibrium reaction.

But I'm still gonna show you how it's solved. When we use Kb, we hydrate ammonia. The equilibrium reaction is

NH₃ + H₂O ↔ NH₄⁺ + OH⁻

Now, we use the ICE(Initial-Change-Equilibrium) analysisL

                     NH₃ + H₂O ↔ NH₄⁺ + OH⁻

Initial             0.55     -             0         0
Change           -x        -             +x      +x
--------------------------------------------------
Equilibrium   0.55 - x                 x        x

The variable x here denotes the moles of the substances that is involved in the reaction. They are balanced out by their stoichiometric ratio which is equal to 1:1. Then, we apply the equation for equilibrium constants

Kb = [NH₄⁺][OH⁻]/[NH₃]

the concentration of water is not included because the solution is very dilute. Substituting the equilibrium amounts:

Kb = [x][x]/[0.55-x]

Since we are given the pH, we can use the relationship
pH = 14 - pOH    ;  pOH = -log [OH⁻]
11.5 = 14 - pOH
pOH = 2.5

2,5 = -log [OH⁻]
[OH⁻]=[x] = 3.162×10⁻³

Thus,
Kb = [3.162×10⁻³][3.162×10⁻³]/]0.55- 3.162×10⁻³]
Kb = 1.82 × 10⁻⁵

The equilibrium constant for ammonia[tex]\left({{{\text{K}}_{\text{b}}}}\right)[/tex] is[tex]\boxed{1.82\times{{10}^{-5}}}[/tex].

Further explanation:

Chemical equilibrium is the state in which the concentration of reactants and products become constant and do not change with time. This is because the rate of forward and backward direction becomes equal. The general equilibrium reaction is as follows:

[tex]{\text{A(g)}}+{\text{B(g)}}\rightleftharpoons{\text{C(g)}}+{\text{D(g)}}[/tex]

Equilibrium constant is the constant that relates the concentration of product and reactant at equilibrium. The formula to calculate the equilibrium constant for general reaction is as follows:

[tex]{\text{K}}=\frac{{\left[{\text{D}}\right]\left[{\text{C}}\right]}}{{\left[{\text{A}}\right]\left[{\text{B}}\right]}}[/tex]

Here, K is the equilibrium constant.

The equilibrium constant for the dissociation of acid is known as [tex]{{\text{K}}_{\text{a}}}[/tex]and equilibrium constant for the dissociation of base is known as[tex]{{\text{K}}_{\text{b}}}[/tex].

The expression that relates pH and pOH is given as follows:

[tex]{\text{pH}}+{\text{pOH}}=14[/tex]                                   …… (1)

Rearrange equation (1) to calculate pOH.

[tex]{\text{pOH}}=14-{\text{pH}}[/tex]                                 …… (2)

Substitute 11.50 for the value of pH in equation (2).

[tex]\begin{aligned}{\text{pOH}}&=14-{\text{11}}{\text{.50}}\\&={\text{2}}{\text{.5}}\\\end{aligned}[/tex]

pOH is the measure of hydroxide ion concentration. The formula to calculate pOH is as follows:

[tex]{\text{pOH}}=-\log\left[{{\text{O}}{{\text{H}}^-}}\right][/tex]                              …… (3)

Here,

[tex]\left[{{\text{O}}{{\text{H}}^-}}\right][/tex] is the concentration of hydroxide ion.

Rearrange equation (3) to calculate [tex]\left[{{\text{O}}{{\text{H}}^-}}\right][/tex].

[tex]\left[{{\text{O}}{{\text{H}}^-}}\right]={10^{-{\text{pOH}}}}[/tex]                              …… (4)

Substitute 2.5 for pOH in equation (4).

[tex]\begin{aligned}\left[{{\text{O}}{{\text{H}}^-}}\right]&={10^{-2.5}}\\&=0.0031622\\\end{aligned}[/tex]

The given equilibrium reaction is,

[tex]{\text{N}}{{\text{H}}_{\text{3}}}+{{\text{H}}_2}{\text{O}}\rightleftharpoons{\text{NH}}_4^++{\text{O}}{{\text{H}}^-}[/tex]

The expression of [tex]{{\text{K}}_{\text{b}}}[/tex]for the above reaction is as follows:

[tex]{{\text{K}}_{\text{b}}}=\frac{{\left[{{\text{NH}}_4^+}\right]\left[{{\text{O}}{{\text{H}}^-}}\right]}}{{\left[{{\text{N}}{{\text{H}}_3}}\right]}}[/tex]                                  …... (5)

The equilibrium concentration of both [tex]{\text{NH}}_4^+[/tex] and [tex]{\text{O}}{{\text{H}}^-}[/tex] is the same.

0.0031622 M of [tex]{\text{O}}{{\text{H}}^-}[/tex]is present at equilibrium so 0.0031622 M out of 0.55 M of  has reacted.

The initial concentration of the aqueous solution is 0.55 M. So the concentration of [tex]{\text{N}}{{\text{H}}_3}[/tex] left at equilibrium is calculated as follows:

[tex]\begin{aligned}\left[{{\text{N}}{{\text{H}}_3}}\right]&={\text{Initial concentration of N}}{{\text{H}}_{\text{3}}}-{\text{Reacted concentration of N}}{{\text{H}}_{\text{3}}}\\&={\text{0}}{\text{.55 M}}-{\text{0}}{\text{.0031622 M}}\\&={\text{0}}{\text{.5468378 M}}\\\end{aligned}[/tex]

The value of[tex]\left[{{\text{O}}{{\text{H}}^-}}\right][/tex]is 0.0031622 M.

The value of [tex]\left[{{\text{N}}{{\text{H}}_3}}\right][/tex] is 0.5468378 M.

The value of [tex]\left[{{\text{NH}}_4^+}\right][/tex] is 0.0031622 M.

Substitute these values in equation (5).

[tex]\begin{aligned}{{\text{K}}_{\text{b}}}&=\frac{{\left({0.0031622\;{\text{M}}}\right)\left({0.0031622\;{\text{M}}}\right)}}{{\left({0.546837{\text{8 M}}}\right)}}\\&=1.82861\times{10^{-5}}\\&\approx1.82\times{10^{-5}}\\\end{aligned}[/tex]

Therefore, equilibrium constant for ammonia is[tex]{\mathbf{1}}{\mathbf{.82\times1}}{{\mathbf{0}}^{{\mathbf{-5}}}}[/tex].

Learn more:

1. Calculation of equilibrium constant of pure water at 25°c: https://brainly.com/question/3467841

2. Complete equation for the dissociation of [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex](aq): https://brainly.com/question/5425813

Answer details:

Grade: Senior school

Subject: Chemistry

Chapter: Equilibrium

Keywords: NH3, OH-, NH4+, H2O, equilibrium, kb, pH, pOH, 14, 11.5, 2.5, aqueous solution, 0.0031622 M, 0.5468378 M.

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