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Temka [501]
3 years ago
8

The gas in the piston is at constant temperature. A student increases the pressure on the piston from 2 atm to 3 atm. The observ

ation will be summarized in a row of the incomplete table below. mc026-1.jpg Row Name Observation Constants 1 Boyle's law ? ? 2 Charles’s law ? ? 3 Gay-Lussac’s law ? Volume, moles of gas 4 Combined gas law ? ? Which row of the table and which observation would the student fill? row 1, volume decreases when temperature decreases row 1, volume decreases when pressure increases row 3, volume decreases when temperature decreases row 3, volume decreases when pressure increases
Chemistry
2 answers:
blagie [28]3 years ago
6 0

Answer:

Option B) Row 1, Volume decreases when pressure increases.

Explanation:

azamat3 years ago
5 0
Answer : Option B) Row 1, Volume decreases when pressure increases.

Explanation : According to the boyle's law, when the pressure  of the gas molecules are increased from  2 atm to  3 atm creates a decrease in volume of the gas which then increases the rate of collision of the molecules and makes it more compact than it was before.
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Calculate the enthalpy change, ∆H in kJ, for the reaction H2O(s) → H2(g) + 1/2O2(g). Use the following information: : +279.9 kJ
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Answer:

+ 291.9 kJ

Solution:

The equation given is as;

H₂O ₍s₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = ?

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H₂ ₍g₎ + 1/2 O₂ ₍g₎ → H₂O ₍l₎ ΔH = - 285.9 kJ

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H₂O ₍l₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 285.9 kJ

Also, we know that,

H₂O ₍s₎ → H₂O ₍l) ΔH = + 6.0 kJ

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H₂O ₍l₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 285.9 kJ

H₂O ₍s₎ → H₂O ₍l) ΔH = + 6.0 kJ

-----------------------------------------------------------------------------

H₂O ₍s₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 291.9 kJ

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Show, in a step by step fashion, your calculations to determine the percent
marusya05 [52]

The percent  by volume of a solution : 19.23%

<h3>Further explanation</h3>

Given

50 ml of ethanol

210  ml of water

Required

The percent  by volume of a solution

Solution

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\tt \%solute(v/v)=\dfrac{solute~volume}{solution~volume}\times 100\%

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