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SSSSS
1 month ago
15

The volume of a gas at 6.0 atm is 2.5 L. What is the volume of the gas at 7.5 atm at the same temperature?

Chemistry
1 answer:
castortr0y [3K]1 month ago
7 0

Greetings!

The result is:

The new volume is: 2L

Rationale:

Because the temperature remains constant, we can apply Boyle's Law to solve this issue.

Boyle's Law stipulates that:

P_{1}V_{1}=P_{2}V_{2}

Where,

P is the gas's pressure.

V is the gas's volume.

According to the information provided:

V_{1}=2.5L\\P_{1}=6.0atm\\P_{2}=7.5atm

Let's put the values into the equation:

2.5L*6.0atm=7.5atm*V_{2}

2.5L*6.0atm=7.5atm*V_{2}\\\\V_{2}=\frac{2.5L*6.0atm}{7.5atm}=\frac{15L.atm}{7.5atm}=2L

Consequently, the new volume is: 2L

Wishing you a lovely day!

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5 grams of NaHCO3 was added to 50 cm3 of 1.5 mol dm-3 hydrochloric acid, causing the temperature to decrease by 6.8 K. What is t
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The balanced chemical equation for the neutralization of HCl with NaHCO_{3} is:

HCl (aq) + NaHCO_{3}(aq)--> NaCl (aq) + H_{2}O(l)+CO_{2} (g)

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Moles of NaHCO_{3} = 5 g *\frac{1 mol}{84 g} = 0.05952 mol NaHCO_{3}

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Assuming the density of the solution is 1.0 g/mL

Mass of HCl solution = 50 g

Overall mass of the solution = 50 g + 5 g = 55 g

To find the heat of neutralization, we calculate:

Q = m C ΔT

where m equals the mass of the solution = 55 g

C represents the specific heat capacity of the solution = 4.184\frac{J}{g. ^{0}C}

ΔT signifies the temperature change = 6.8 K = (6.8 - 273) C = -266.2^{0}C

Q = 55 g * 4.184 \frac{J}{g. K}(6.8K) = 1565 J

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2 months ago
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Write the balanced molecular and net ionic equation for the reaction that occurs when the contents of the two beakers are added
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