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blsea [12.9K]
3 years ago
15

What is the volume of 0.23kg of pure water

Chemistry
2 answers:
tatyana61 [14]3 years ago
8 0

<u>Answer:</u> The volume of pure water is 230cm^3

<u>Explanation:</u>

To calculate density of a substance, we use the equation:

Density=\frac{Mass}{Volume}

We are given:

Density of pure water = 1g/cm^3

Mass of pure water = 0.23kg = 230 g     (Conversion factor: 1 kg = 1000 g)

Putting values in above equation, we get:

1g/cm^3=\frac{230g}{\text{Volume of pure water}}\\\\\text{Volume of pure water}=230cm^3

Hence, the volume of pure water is 230cm^3

vichka [17]3 years ago
7 0
M = 0,23kg = 230g
d = 1g/cm³

V = 230g / 1g/cm³ = 230cm³ = 0,23L
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Calculate the ΔG°rxn using the following information at 298K. 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) ΔG°rxn = ? ΔH°f (kJ/mol) -2
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Answer:

ΔG°rxn = +50.8 kJ/mol

Explanation:

It is possible to obtain ΔG°rxn of a reaction at certain temperature from ΔH°rxn and S°rxn, thus:

<em>ΔG°rxn = ΔH°rxn - T×S°rxn (1)</em>

In the reaction:

2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l)

ΔH°rxn = 3×ΔHfNO2 + ΔHfH2O - (2×ΔHfHNO3 + ΔHfNO)

ΔH°rxn = 3×33.2kJ/mol + (-285.8kJ/mol) - (2×-207.0kJ/mol + 91.3kJ/mol)}

ΔH°rxn = 136.5kJ/mol

And S°:

S°rxn = 3×S°NO2 + S°H2O - (2×S°HNO3 + S°NO)

ΔH°rxn = 3×0.2401kJ/molK + (0.0700kJ/molK) - (2×0.146kJ/molK + 0.2108kJ/molK)

ΔH°rxn = 0.2875kJ/molK

And replacing in (1) at 298K:

ΔG°rxn = 136.5kJ/mol - 298K×0.2875kJ/molK

<em>ΔG°rxn = +50.8 kJ/mol</em>

<em />

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The cranial region encompasses the upper part of the head.

∴ The first answer choice matches its description.

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The axillary region is an anatomical region under the shoulder joint where the arm connects to the shoulder. Therefore, it encompasses the armpits.

∴ The second answer choice matches its description.

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The thoracic region runs from the base of the neck down to the abdomen. Therefore, it encompasses the chest

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