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3 years ago

(NH4)3PO4 + Pb(NO3)4. Pb3(PO4)4 + NH4NO3

Chemistry

1 answer:

Dafna1 [17]3 years ago

6 0

4(NH₄)₃PO₄ + 3Pb(NO₃)₄ → Pb₃(PO₄)₄ + 12NH₄NO₃

Explanation:

(NH₄)₃PO₄ + Pb(NO₃)₄ → Pb₃(PO₄)₄ + NH₄NO₃

Balancing this equation;

We simply use a mathematical approach by solving simple algebraic equations.

a(NH₄)₃PO₄ + bPb(NO₃)₄ → cPb₃(PO₄)₄ + dNH₄NO₃

a,b,c and d are the coefficients that will balance the equation;

Conserving N: 3a + 4b = 2d equation 1

H: 12a = 4d

P; a = 4c

O; 4a + 12b = 16c + 3d

Let us make a = 1;

c = $\frac{1}{4}$

d = 3

solving for b;

from (i); 4b = 2d -3a = 2(3) - 3(1) = 3

b = \frac{3}{4}

Now multiply a, b, c and d by 4;

a = 4 , b = 3 , c= 1 and d = 12

4(NH₄)₃PO₄ + 3Pb(NO₃)₄ → Pb₃(PO₄)₄ + 12NH₄NO₃

Learn more:

Balanced equation brainly.com/question/11102790

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The reaction between zinc (Zn) and hydrogen chloride (HCl) produces zinc chloride (ZnCl), hydrogen (H2) gas, and heat. If more h

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If more heat is removed from the reaction the rate of reaction change as below to counter the action

The rate of the forward reaction increase and produces more zinc chloride

explanation

The reaction of zinc and HCl to produce ZnCl and H2 is exothermic reaction, heat is produced as one product and by removing heat it favor forward reaction

The position of equilibrium moves to the right since removing heat led to decrease of temperature and more zinc chloride is produced.

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3 years ago

Consider the following reaction:

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For the given reaction, according to the Law of Conservation of Energy, the energy required to decompose Hcl and produce $H_{2}+c l_{2}$ are equal.

Answer: Option C

Explanation:

According to law of conservation's of energy, energy can only be transferred from reactants to product side. So in this process, it is stated that 185 kJ of energy will be needed to decompose it. So that 185 kJ of energy will be getting transferred to produce the creation of hydrogen and chloride in the product side.

So if we see from the reactants side, the energy of 185 kJ is required for decomposition of hydrogen chloride. Similarly, if we see from the product side, the 185 kJ utilized for decomposition is transferred as energy required to create hydrogen and chlorine atoms. This statement will be in accordance with the law of conservation's of energy.

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4 years ago

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What did you include in your response? Check all that apply.

Aliun [14]

Answer:

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Calculate the percentage difference in the fundamental vibrational wavenumbers of 23Na35Cl and 23Na37Cl on the assumption that t

stealth61 [152]

Answer:

1.089%

Explanation:

From;

ν =1/2πc(k/meff)^1/2

Where;

ν = wave number

meff = reduced mass or effective mass

k = force constant

c= speed of light

Let

ν =1/2πc (k/meff)^1/2 vibrational wave number for 23Na35 Cl

ν' =1/2πc(k'/m'eff)^1/2 vibrational wave number for 23Na37 Cl

The between the two is obtained from;

ν' - ν /ν = (k'/m'eff)^1/2 - (k/meff)^1/2 / (k/meff)^1/2

Therefore;

ν' - ν /ν = [meff/m'eff]^1/2 - 1

Substituting values, we have;

ν' - ν /ν = [(22.9898 * 34.9688/22.9898 + 34.9688) * (22.9898 + 36.9651/22.9898 * 36.9651)]^1/2 -1

ν' - ν /ν = -0.01089

percentage difference in the fundamental vibrational wavenumbers of 23Na35Cl and 23Na37Cl;

ν' - ν /ν * 100

|(-0.01089)| × 100 = 1.089%

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3 years ago

Which of these half-reactions represents reduction?

gogolik [260]

Answer: The half-reactions represents reduction are as follows.

$Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}$
$MnO^{-}_{4} \rightarrow Mn^{2+}$

Explanation:

A half-reaction where addition of electrons take place or a reaction where decrease in oxidation state of an element takes place is called reduction-half reaction.

For example, the oxidation state of Cr in $Cr_{2}O^{2-}_{7}$ is +6 which is getting converted into +3, that is, decrease in oxidation state is taking place as follows.

$Cr_{2}O^{2-}_{7} + 3 e^{-} \rightarrow Cr^{3+}$

Similarly, oxidation state of Mn in $MnO^{-}_{4}$ is +7 which is getting converted into +2, that is, decrease in oxidation state is taking place as follows.

$MnO^{2-}_{4} + 5 e^{-} \rightarrow Mn^{2+}$

Thus, we can conclude that half-reactions represents reduction are as follows.

$Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}$
$MnO^{-}_{4} \rightarrow Mn^{2+}$

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3 years ago