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Pavel [41]
3 years ago
7

5. How many moles are in 58.6 L of Nitrogen gas (N.) at STP?

Chemistry
1 answer:
pantera1 [17]3 years ago
6 0

Answer:

2.6166 moles

Explanation:

stp means: T=273k   p=101.3kpa

R= 8.31(constant)       v=58.6  

using the ideal gas law; pv=nRt

                                        101.3 x 58.6 = n x 8.31 x 273

                                         n= (101.3 x 58.6)\(8.31 x 273)

                                         n= 2.6166moles

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True or false: the most easily ionizable elements are the most electronegative.
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3 0
2 years ago
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
Completely reacting 150.0 g of a substance with oxygen releases 395.1 J of energy. How much energy would be released if 450.0 g
MrMuchimi
To solve this problem we just need to use the rule of three:
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x = 450*395.1/150 = 1185,3J

450.0 g of the substance completely reacted with oxygen will produce 1.1853 kJ(<span>kiloJoule</span>)
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Answer:

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