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tatuchka [14]
3 years ago
6

A sample of 35.1 g of methane gas has a volume of 5.20 l at a pressure of 2.70 atm. calculate the temperature of the gas.

Chemistry
1 answer:
melisa1 [442]3 years ago
6 0
We can use the ideal gas equation to determine the temperature with the given conditions of mass of the gas, volume, and pressure. The equation is expressed
PV=nRT where n is the number of moles equal to mass / molar mass of gas. Substituting the given conditions with R = 0.0521 L atm/mol K we can find the temperature
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Which describes the current model of the atom? (2 points)
Harman [31]
A. I hope this helps!
4 0
3 years ago
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Is the answer B? Help
Masja [62]

Answer:

A

Explanation:

Hmm, so we have the following in the diagram

Pt(s)

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Pt 2+, 4+, 6+  Though it states Pt is inert

Cl 2-

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Anode definition: the positively charged electrode by which the electrons leave an electrical device.

Electrode definition: a conductor through which electricity enters or leaves an object, substance, or region.

Cations attracted to cathode pick up electrons

Anions attracted to anode release electrodes+

Reduction at Cathode (red cat gain of e)

Oxidation at Anode (ox anode loss of e)

So from the diagram we can see that the charge is being generated through the 2 metal plates.

So the answer is A, the anode material is Pt and the half reaction is 2Cl- = Cl2 + 2e-

7 0
2 years ago
How do you convert between the mass and the number of moles of a substance?
Ksivusya [100]
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3 years ago
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7. NH2CO2NH4(s) when heated to 450 K undergoes the following reaction to produce a system which reaches equilibrium: NH2CO2NH4(s
Taya2010 [7]

Answer:

Value of equilibrium constant is 0.0888

Explanation:

Both NH_{3} and CO_{2} are gaseous. Hence equilibrium constant depends upon partial pressures of NH_{3} and CO_{2}.

Initially no NH_{3} and CO_{2} were present.

Hence mole fraction of NH_{3} and CO_{2} at equilibrium can be calculated from coefficient of NH_{3} and CO_{2} in balanced equation.

Mole fraction of NH_{3} = (number of moles of NH_{3})/(total number of moles of NH_{3} and CO_{2}) = \frac{2moles}{(2+1)moles}=\frac{2}{3}

Mole fraction of CO_{2} = (number of moles of CO_{2})/(total number of moles of NH_{3} and CO_{2}) = \frac{1moles}{(2+1)moles}=\frac{1}{3}

Let's assume both CO_{2} and NH_{3} behaves ideally.

Therefore partial pressure of NH_{3}, P_{NH_{3}}= x_{NH_{3}}.P_{total} and P_{CO_{2}}= x_{CO_{2}}.P_{total}

Where x represents mole fraction

So, P_{NH_{3}}=\frac{2}{3}\times 0.843atm=0.562atm

P_{CO_{2}}=\frac{1}{3}\times 0.843atm=0.281atm

So, K_{p}=P_{NH_{3}}^{2}.P_{CO_{2}}=(0.562)^{2}\times 0.281=0.0888

4 0
3 years ago
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