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3 years ago

Which statement is true regarding methane and ammonia?

2 answers:

6 0

The options for given question are as follow,

1) Methane molecules show hydrogen bonding.
<span>2) Ammonia molecules show hydrogen bonding. </span>
<span>3) Methane has stronger hydrogen bonding than ammonia. </span>
<span>4) Both the compounds do not show hydrogen bonding. </span>
<span>5) Both the compounds have strong hydrogen bonding.
</span>
Answer:
Correct answer is Option-2 (Ammonia molecules show hydrogen bonding).

Explanation:
Hydrogen bond interactions are formed when a partial positive hydrogen atom attached to most electronegative atom of one molecule interacts with the partial negative most electronegative element of another molecule. So, in Ammonia hydrogen gets partial positive charge as nitrogen is highly electronegative. While the C-H bond in Methane is non-polar and fails to form hydrogen bond interactions.

Olenka [21]3 years ago

4 0

The answer is B. Trust me!!!

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The partial pressure of CO2 gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO2 gas (in g) will be released f

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If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. We can calculate the concentration of CO₂ using Henry's law.

$C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 4.60 atm = 7.59 \times 10^{-3} M$

We can calculate the mass of CO₂ in 1.1 L considering its molar mass is 44.01 g/mol.

$\frac{7.59 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.367 g$

Now, we will repeat the same procedure for a partial pressure of 1.28 atm.

$C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 1.28 atm = 2.11 \times 10^{-3} M$

$\frac{2.11 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.102 g$

The mass of CO₂ released will be equal to the difference in the masses at the different pressures.

$m = 0.367 g - 0.102 g = 0.265 g$

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

Learn more: brainly.com/question/18987224

<em>The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO₂ gas (in g) will be released from 1.1 L of the carbonated water when the partial pressure of CO2 is lowered to 1.28 atm? At 25 ºC, the Henry’s law constant for CO₂ dissolved in water is 1.65 x 10⁻³ M/atm, and the density of water is 1.0 g/cm³.</em>

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. the density (d) of a substance is an intensive property that is defined as the ratio of its mass (m) to its volume (v). densit

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Density is intensive because it is the ratio of two extensive properties that is mass to volume.

<h3>What is Density ?</h3>

Density is defined as mass per unit volume. S.I unit if density is kg/m³.

It is expressed as

Density = $\frac{\text{Mass}}{\text{Volume}}$ or $d = \frac{m}{V}$

<h3>What is Intensive Property ? </h3>

Intensive property is the property which does not depend on the size of the system. Intensive property can be easily identified. Intensive property is the independent property. The size does not change in this.

Example of intensive property are Density, Freezing point, melting point, colour, Lustre, etc.

<h3>What is Extensive property ?</h3>

Extensive property is the property which depend on the substance. Extensive property cannot be easily identified.

Example of extensive property length, weight, volume, mass etc.

Thus from the above conclusion we can say that Density is intensive because it is the ratio of two extensive properties that is mass to volume.

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Disclaimer: The given question is incomplete on the portal. Here is the complete question.

Question: The density (d) of a substance is an intensive property that is defined as the ratio of its mass (m) to its volume (v)

density = $\frac{\text{Mass}}{\text{Volume}}$ or $d = \frac{m}{V}$

Considering that mass and volume are both extensive properties, explain why their ratio, density is intensive.

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