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Vlad1618 [11]
4 years ago
5

Calculate the change in enthalpy for the following reaction, using the bond enthalpy data provided below. (Note: you may round t

he bond energies to the nearest 100 kJ/mol during your calculations.) 2 N2H2(g) 3 O2(g)
Chemistry
1 answer:
aliina [53]4 years ago
3 0

Answer:

-514 kJ/mol

Explanation:

The bond enthalpy which is also known as bond energy can be defined as the amount of energy needed to split one mole of the stated bond. The change in enthalpy of a given reaction can be estimated by subtracting the sum of the bond energies of the reactants from the sum of the bond energies of the products.

For the given chemical reaction, the change in enthalpy of the reaction is:

ΔH_{rxn} = [2(409) + 4(388) + 3(496) - 4(630) - 4(463)] kJ/mol = 818 + 1552 + 1488 - 2520 - 1852 = -514 kJ/mol

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Answer:

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Explanation:

<em>The volume of the glucose solution to be prepared</em> = 500 cm^3

<em>Molarity of the glucose solution to be prepared</em> = 1 M

i. Molar mass of glucose (C_1_2H_6O_6) = (6 × 12) + (12 × 1) + (6 × 16) = 180 g/mol

ii.<em> mole = molarity x volume</em>. Hence;

amount (in moles) of the glucose solution to be prepared

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3 0
3 years ago
What type of bond would form between two atoms of selenium?
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A researcher wants to experiment with an element that reacts like phosphorus (P) but has a greater atomic mass. Which element sh
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4 0
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Read 2 more answers
write a correct and complete nuclear equation for the alpha decay of a parent atom having 102 protons and 167 neutrons useing th
scZoUnD [109]
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after an alpha decay, the new element formed has 100 protons which is Fm ( Fermium)

the alpha decaying equation is,

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5 0
3 years ago
Lead ions can be precipitated from aqueous solutions by the addition of aqueous iodide: Pb2+ (aq) + 2I− (aq) → PbI2 (s) Lead iod
Angelina_Jolie [31]

Answer:

338.00 mL

Explanation:

The lead ions come from the salt Pb(NO₃)₂ and the iodide from the acid HI, so the balanced reaction is:

Pb(NO₃)₂(aq) + 2HI(aq) → PbI₂(s) + 2HNO₃(aq)

So, the stoichiometry is 1 mol of Pb(NO₃)₂ to 2 moles of HI, then:

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V = 1.200/3.550 = 0.338 L

V = 338.00 mL

6 0
3 years ago
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