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Fantom [35]
3 years ago
8

Write the complete balanced equation for the following reaction: Na2SO4 + Pb(NO3)2 -->NaNO3 + PbSO4

Chemistry
2 answers:
emmasim [6.3K]3 years ago
7 0

Explanation:

A reaction equation that shows same number of atoms on both reactant and product sides is known as a balance equation.

Na_{2}SO_{4} + Pb(NO_{3})_{2} \rightarrow NaNO_{3} + PbSO_{4}

Number of atoms on reactant side are as follows.

Na = 2

SO_{4} = 1

Pb = 1

NO_{3} = 2

Number of atoms on product side are as follows.

Na = 1

SO_{4} = 1

Pb = 1

NO_{3} = 1

Therefore, to balance this chemical equation we multiply NaNO_{3} by 2 on product side. Thus, we can conclude that the balanced chemical equation will be as follows.

            Na_{2}SO_{4} + Pb(NO_{3})_{2} \rightarrow 2NaNO_{3} + PbSO_{4}

Lyrx [107]3 years ago
6 0
Na2SO4 + Pb(NO3)2 --> NaNO3 + PbSO4
Na2SO4 + Pb(NO3)2 --> 2NaNO3 + PbSO4

(i'm not sure if you need to add the forms of the compounds as well)
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adoni [48]

Answer: .B. 13C NMR spectra display peaks for only carbons that bear hydrogen atoms.

Explanation:

The statements that are true about 13C NMR are:

A. In 13C proton-decoupled NMR spectra, all peaks are singlets.

C 13C NMR chemical shifts occur over a greater range than 1H NMR chemical shifts.

D. 13C NMR easily differentiates between the different hybridized carbons (sp3, sp2, and sp hybridized carbons).organic-chemistry

Therefore, the option that isn't true is option B. "13C NMR spectra display peaks for only carbons that bear hydrogen atoms". This is false because 13C NMR will show every peak in the spectrum and it doesn't matter if it's only carbons that bear hydrogen atoms as everything will be shown.

4 0
3 years ago
Phosphoglucoisomerase interconverts glucose 6‑phosphate to, and from, glucose 1‑phosphate.
lawyer [7]

Answer:

Keq = 0.053

7.3 kJ/mol

Explanation:

Let's consider the following isomerization reaction.

glucose 6‑phosphate ⇄ glucose 1 - phosphate

The concentrations at equilibrium are:

[G6P] = 0.19 M

[G1P] = 0.01 M

The concentration equilibrium constant (Keq) is:

Keq = [G1P] / [G6P]

Keq = 0.01 / 0.19

Keq = 0.053

We can find the standard free energy change, ΔG°, of the reaction mixture using the following expression.

ΔG° = -R × T × lnKeq

ΔG° = -8.314 J/mol.K × 298 K × ln0.053

ΔG° = 7.3 × 10³ J/mol = 7.3 kJ/mol

7 0
4 years ago
At 298 K, what is the Gibbs free energy change (ΔG) for the following reaction?
9966 [12]

Answer:

(a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

Explanation:

Given that,

Temperature = 298 K

Suppose, density of graphite is 2.25 g/cm³ and density of diamond is 3.51 g/cm³.

\Delta H\ for\ diamond = 1.897 kJ/mol

\Delta H\ for\ graphite = 0 kJ/mol

\Delta S\ for\ diamond = 2.38 J/(K mol)

\Delta S\ for\ graphite = 5.73 J/(K mol)

(a) We need to calculate the value of \Delta G for diamond

Using formula of Gibbs free energy change

\Delta G=\Delta H-T\Delta S

Put the value into the formula

\Delta G= (1897-0)-298\times(2.38-5.73)

\Delta G=2895.3

\Delta G=2.895\ kJ

The Gibbs free energy  change is positive.

(b). When it is compressed isothermally from 1 atm to 1000 atm

We need to calculate the change of Gibbs free energy of diamond

Using formula of gibbs free energy

\Delta S=V\times\Delta P

\Delta S=\dfrac{m}{\rho}\times\Delta P

Put the value into the formula

\Delta S=\dfrac{12\times10^{-6}}{3.51}\times999\times10130

\Delta S=34.59\ J/mole

(c). Assuming that graphite and diamond are incompressible

We need to calculate the pressure

Using formula of Gibbs free energy

\beta= \Delta G_{g}+\Delta G+\Delta G_{d}

\beta=V(-\Delta P_{g})+\Delta G+V\Delta P_{d}

\beta=\Delta P(V_{d}-V_{g})+\Delta G

Put the value into the formula

0=\Delta P(\dfrac{12\times10^{-6}}{3.51}-\dfrac{12\times10^{-6}}{2.25})\times10130+2895.3

0=-0.0194\Delta P+2895.3

\Delta P=\dfrac{2895.3}{0.0194}

\Delta P=14924\ atm

(d). Here, C_{p}=0

So, The value of \Delta H and \Delta S at 900 k will be equal at 298 K

We need to calculate the Gibbs free energy of diamond relative to graphite

Using formula of Gibbs free energy

\Delta G=\Delta H-T\Delta S

Put the value into the formula

\Delta G=(1897-0)-900\times(2.38-5.73)

\Delta G=4912\ J

Hence, (a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

7 0
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Lead nitrate can be decomposed by heating. What is the percent yield of the decomposition reaction if 9.9 g of Pb(NO3)2 are heat
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w-?

w=m'{PbO}/m{PbO}

m{PbO}=M{PbO}m{Pb(NO₃)₂}/M{Pb(NO₃)₂}

w=m'{PbO}M{Pb(NO₃)₂}/[M{PbO}m{Pb(NO₃)₂}]

w=5.5*331.2/[223.2*9.9]=0.8244 (82.44%)
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Explanation:

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