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Volgvan
3 years ago
13

Which equivalent factor should you use to convert from 4.28 * 10^19 molecules of oxygen to moles of O2

Chemistry
1 answer:
stiv31 [10]3 years ago
5 0
The equivalent factor used to convert molecules of oxygen to moles of oxygen is Avogadro's number which is 6.022 x10^23. We divide the given number of molecules by Avogadro's number to get the number of moles. In this case,  4.28 * 10^19 molecules divided by <span>6.022 x10^23 is equal to 7.11 x10^-5 moles.</span>
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There are some exceptions to the trends of first and successive ionization energies. For each of the following pairs, explain wh
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Answer:

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Read 2 more answers
I want to know the steps.
Artyom0805 [142]

The answer for the following problem is described below.

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

Explanation:

Given:

enthalpy of combustion of glucose(ΔH_{f} of C_{6}H_{12} O_{6}) =-1275.0

enthalpy of combustion of oxygen(ΔH_{f} of O_{2}) = zero

enthalpy of combustion of carbon dioxide(ΔH_{f} of CO_{2}) = -393.5

enthalpy of combustion of water(ΔH_{f} of H_{2} O) = -285.8

To solve :

standard enthalpy of combustion

We know;

ΔH_{f}  = ∈ΔH_{f} (products) - ∈ΔH_{f} (reactants)

C_{6}H_{12} O_{6} (s) +6 O_{2}(g) → 6 CO_{2} (g)+ 6 H_{2} O(l)

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [0 - 1275]

ΔH_{f} = 6 (-393.5) + 6(-285.8)  - 0 + 1275

ΔH_{f} = -2361 - 1714 - 0 + 1275

ΔH_{f} =-2800 kJ

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

7 0
3 years ago
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