What is the percentage of nitrogen in the fertilizer urea ((nh2) 2co)

Chemistry

1 answer:

maks197457 [2]3 years ago

8 0

Answer:

You have 1680 grams of urea. It means (in it) there are 787 grams of nitrogen, 112 grams of hydrogen, 336 grams of carbon, and 448 grams of oxygen.

Explanation:

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Calculate the cell potential, E, for the following reactions at 26.29 °C using the ion concentrations provided. Then, determine

yanalaym [24]

Answer:

I will work only one of the listed equations ... you follow the given example for the remaining reactions. Thank you :-)

Rxn 1: Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s)

a) E(Pt⁺²/Fe°) = - 1.668v

b) Process is Non-spontaneous if E(cell) < 0

Explanation:

Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s) ⇔

Pt°(s)|Pt⁺²[0.057M]║Fe⁺²[0.006M]|Fe°(s)

As written, Pt° is shown undergoing oxidation with Fe⁺² undergoing reduction. Applying the reduction potentials to the analytical equations for E(cell) and ΔG(cell) gives E(Pt/Fe⁺²) < 0 and ΔG(Pt/Fe⁺²) > 0 which indicate a non-spontaneous process. The following supports this conclusion.

E°(Fe⁺²) = -0.44v

E°(Pt⁺²) = +1.20v

E°(Pt/Fe⁺²) =E°(Redn) - E°(Oxidn) =E°(Fe⁺²) - E°(Pt⁺²)

= -0.44v - (+1.20v) = - 1.64v

[Fe⁺²] = 0.0066M

[Pt⁺²] = 0.057M

n = electrons transferred = 2

E(nonstd) = E°(std) - (0.0592/n)logQ);

Q = [Pt⁺²]/[Fe⁺²]

= -1.64v - (0.0592/2)log[0.057M]/[0.006M]v = -1.668v

Also, if ΔG(cell) > 0 => indicates non-spontaneous process

ΔG(Pt/Fe⁺²) = - nFE = -(2)(96,500Coulombs)((-1.664v) > 0 Kj => nonspontaneous rxn. (1 Coulomb-volt = 1 Kilojoule)

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3 years ago

What is the gram-formula mass of Ca3(PO4)2

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310.17 grams so there

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3 years ago

Clo3− draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons. Show t

Karo-lina-s [1.5K]

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Solution:-

<u>O 7-4 = 3 O Double bond on one H 5-4 = 1</u>

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O 3-4= -1 O O 6-7 = -1(2)=-2 H 1-0=+1

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Learn more about Nonhydrogen atoms here:-brainly.com/question/2822744

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