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2 years ago

PLEASE HELP ASAP!!

1 answer:

3 0

At STP (Standard Temperature and Pressure) there are 22.4 Liters in 1 mole.
Also there is 4g per every mole of Helium. You will use dimensional analysis to find the answer. You first start with grams, convert to moles, then convert moles to liters like so.

He = Helium

7500g He = 1 mole He = 1875 mole He
—————
4g He

1875 moles He = 22.4 L He = 42,000L He
—————
1 mole He

Answer should be 42,000 L of He

(Are you sure the question is 7,500 grams? Or does it say 75.00 grams?)

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In a fixed cylinder are 3moles of oxygen gas at 300Kelvin and 1.25atm. What is the volume of the container?

vladimir2022 [97]

Answer:

The volume of the container is 59.112 L

Explanation:

Given that,

Number of moles of Oxygen, n = 3

Temperature of the gas, T = 300 K

Pressure of the gas, P = 1.25 atm

We need to find the volume of the container. For a gas, we know that,

PV = nRT

V is volume

R is gas constant, R = 0.0821 atm-L/mol-K

So,

$V=\dfrac{nRT}{P}\\\\V=\dfrac{3\ mol\times 0.0821\ L-atm/mol-K \times 300\ K}{1.25\ atm}\\\\V=59.112\ L$

So, the volume of the container is 59.112 L

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2 years ago

Daniel Harris has always enjoyed eating tuna fish. Unfortunately, a study of the mercury content of canned tuna in 2010 found th

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That would be 3 days 6 oz of tuna

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2 years ago

A sample of neon gas at a pressure of 1.08 atm fills a flask with a volume of 250 mL at a temperature of 24.0 °C. If the gas is

musickatia [10]

Answer:

124.91mL

Explanation:

Given parameters:

P₁ = 1.08atm

V₁ = 250mL

T₁ = 24°C

P₂ = 2.25atm

T₂ = 37.2°C

V₂ = ?

Solution:

To solve this problem, we are going to apply the combined gas law;

$\frac{P_{1} V_{1} }{T_{1} } = \frac{P_{2} V_{2} }{T_{2} }$

P, V and T represents pressure, volume and temperature

1 and 2 delineates initial and final states

Convert the temperature to kelvin;

T₁ = 24°C, T₁ = 24 + 273 = 297K

T₂ = 37.2°C , T₂ = 37.2 + 273 = 310.2K

Input the variables and solve for V₂

$\frac{1.08 x 250}{298} = \frac{2.25 x V_{2} }{310.2}$

V₂ = 124.91mL

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2 years ago

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple _______. Then, plot ln(K

Aloiza [94]

Answer:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us use the thermodynamic definition of the Gibbs free energy and its relationship with Ksp as follows:

$\Delta G=-RTln(Ksp)\\\\\Delta G=\Delta H-T\Delta S$

Thus, by combining them, we obtain:

$-RTln(Ksp)=\Delta H-T\Delta S\\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{T\Delta S}{RT} \\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{\Delta S}{R}$

Which is related to the general line equation:

$y=mx+b$

Whereas:

$y=ln(Ksp)\\\\m=-\frac{\Delta H}{R} \\\\x=\frac{1}{T} \\\\b=\frac{\Delta S}{R}$

It means that we answer to the blanks as follows:

Regards!

8 0

2 years ago

For a particular reaction, the change in enthalpy is â9kJmole and the activation energy is 13kJmole. The enthalpy change (ÎH) an

Svetllana [295]

Answer:

The answer is Option a, that is "−9kJmole,5kJmole".

Explanation:

Please find the complete question in the attached file.

In the question, it uses the catalyst inside a process, which does not modify the process eigenvalues, however, it decreases the active energy with an enthalpy of -9kJmole, and also the power for activating decreases around 13 to 5 kJ mole, that's why the choice a is correct.

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2 years ago