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Ann [662]
3 years ago
8

After writing and balancing the equation, what's the next step in using bond energies to calculate the enthalpy of a reaction? Q

uestion 11 options: A) Subtract the bond energies of the products from the reactants. B) Sum the bond energies of the products. C) Write the structures of the products and reactants. D) Write down the bond energies for the products and reactants.
Chemistry
1 answer:
Bas_tet [7]3 years ago
8 0

Answer:

D. write down the bond energies for the reactants and product.

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Balance the following equation. Then determine the ratio for the products KCl and O2 generated during the decomposition of potas
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Consider n2 (g) + 3h2 (g) →→ 2nh3 (g). what is the mass of nitrogen gas required to react with 0.129 g h2?
Flura [38]
The given chemical reaction given above is already balanced such that the number of atoms in the left hand side of the equation is equal to that of the right hand side. Using the dimensional analysis, proper conversion factors and the molar masses,

                    mass of nitrogen = (0.129 g H₂)(1 mol H₂/2 g H₂)(1 mol N₂/3 mol H₂)(28 g N₂/1 mol N₂)
                     mass of nitrogen = 0.602 g N₂

Therefore, 0.602 g of nitrogen will be required for he reaction. 
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3 years ago
Convert 3.52 × 10^22 atoms of gold to moles​
IrinaVladis [17]

Answer:

5.85 x 10⁻² mol Au

Explanation:

Divide the atoms of gold with Avogadro's number to find the solution.

5 0
3 years ago
What quantity of energy would be necessary to boil water with a mass of
anygoal [31]

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3 years ago
How many excess electrons must be added to an isolated spherical conductor 41.0 cmcm in diameter to produce an electric field of
alina1380 [7]

Answer:

3.65 x 10¹⁰ electrons

Explanation:

we'll apply the following equation for electric field of a point charge on a spherical conductor

E = k \frac{q}{r^{2} }

where E is the electric field

k is a constant of the value 8.99 x 10⁹ Nm²/C²

r is the radius of the spherical conductor

q is the total charge in the sphere

Given diameter d =41.0cm, radius r = 20.5cm = 0.205m (convert cm to m)

Electrical field E = 1250 N/C

we are asked to determine how many excess electrons must be added to the surface of the sphere to produce this electric field

E = k \frac{q}{r^{2} }

q = <u>E x r²</u>

        k

q =  <u>1250 N/C x 0.205m</u>²

       8.99 x 10⁹ Nm²/C²

q =   5.84 x 10⁻⁹ C

this is the total charge in the sphere

To determine the number of electrons, we can divide the charge q by the charge on an electron e (1.6 x 10⁻¹⁹C)

n = \frac{q}{e}

n = <u>5.84 x 10⁻⁹ C </u>

       1.6 x 10⁻¹⁹C

n = 3.65 x 10¹⁰ electrons

Therefore, to apply an electric field of magnitude 1250 N/C, the isolated spherical conductor must contain 3.65 x 10¹⁰ electrons

3 0
3 years ago
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