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3 years ago

Difference between qualitative and quantitative data in chemistry

Chemistry

1 answer:

const2013 [10]3 years ago

7 0

Quantitative data is numerical.
Qualitative data is non-numerical.
Hope this helps.
have a great day.

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Write balanced equations for each of the processes described below. (Use the lowest possible coefficients. Omit states-of-matter

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Explanation:

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3 years ago

What is the charge of oxygen in Na2O?

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In Na2O, what is the oxidation state of oxygen? In Na2O oxidation state of Na is 1+

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2 years ago

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or a particular isomer of C 8 H 18 , the combustion reaction produces 5104.1 kJ of heat per mole of C 8 H 18 ( g ) consumed, und

beks73 [17]

Answer:

The standard enthalpy of formation of this isomer of $C_{8}H_{18}$ is -220.1 kJ/mol.

Explanation:

The given chemical reaction is as follows.

$C_{8}H_{18}(g)+ \frac{25}{2}O_{2}(g) \rightarrow 8CO_{2}(g)+9H_{2}O(g)$

$\Delta H^{o}_{rxn}= -5104.1kJ/mol$

The expression for the entropy change for the reaction is as follows.

$\Delta H^{o}_{rxn}=[8\Delta H^{o}_{f}(CO_{2}) +9\Delta H^{o}_{f}(H_{2}O)]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}\Delta H^{o}_{f}(O_{2})]$

$\Delta H^{o}_{f}(H_{2}O)= -241.8kJ/mol$

$\Delta H^{o}_{f}(CO_{2})= -393.5kJ/mol$

$\Delta H^{o}_{f}(O_{2})= 0kJ/mol$

Substitute the all values in the entropy change expression.

$-5104.1kJ/mol=[8(-393.5)+9(-241.8)kJ/mol]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}(0)kJ/mol]$

$-5104.1kJ/mol=-5324.2kJ/mol -\Delta H^{o}_{f}(C_{8}H_{18})$

$\Delta H^{o}_{f}(C_{8}H_{18}) =-5324.2kJ/mol +5104.1kJ/mol$

$=-220.1kJ/mol$

Therefore, The standard enthalpy of formation of this isomer of $C_{8}H_{18}$ is -220.1 kJ/mol.

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3 years ago

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Answer:

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Explanation:

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2 years ago

How many particles are present in 10.0 moles of table salt?

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1.01 x 10^24 molecules.

Explanation:

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10 molecules = 1.68 mol x (6.022 x 10^23 molecules) / (1 mol = 1.01 x 10^24) molecules.

I hope this helps :)

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2 years ago