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JulsSmile [24]
3 years ago
14

What is the pressure of 5.0 mol nitrogen gas in a 2.0 L container at 286 K

Chemistry
1 answer:
g100num [7]3 years ago
5 0
The "Ideal Gas Law Equations" is 
PV=nRT
P= Pressure (in Pascals)
V=Volume (in Liters)
n=amount, or <u>n</u>umber (in moles)
R= 8.3145 \frac{Pascals*Liters}{Moles*Kelvin} or \frac{PaL}{molK}
T= Temperature (In Kelvin)

P= \frac{nRT}{V}

Plug into the equation and you're good!
P=\frac{nRT}{V}
P =\frac{(5mol)(8.3145)(286K)}{2L}
P=5944.8675Pa

If your teacher cares about sig figs,
2 sig figs (significant figures)
P=5900Pa

For other units of pressure,
1 atm = 760 mmHG = 760 Torr = 101326 Pa = 1.01325 bar<u />
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How many milliliters of ammonium sulfate solution having a concentration of 0.218 M are needed to react completely with 62.6 ml
BartSMP [9]

Answer:

330 mL of (NH₄)₂SO₄ are needed

Explanation:

First of all, we determine the reaction:

(NH₄)₂SO₄  +  2NaOH →  2NH₃  +  2H₂O  +  Na₂SO₄

We determine the moles of base:

(First, we convert the volume from mL to L) → 62.6 mL . 1L/1000 mL = 0.0626L

Molarity . volume (L) = 2.31 mol/L . 0.0626 L = 0.144 moles

Ratio is 2:1. Therefore we make a rule of three:

2 moles of hydroxide react with 1 mol of sulfate

Then, 0.144 moles of NaOH must react with (0.144 .1) /2 = 0.072 moles

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6 0
2 years ago
What is the abrviation of the element gold on the periodic table.
Natalka [10]

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Hope this helps Buddy!


- Courtney

6 0
3 years ago
Read 2 more answers
The decomposition of N2O5 in solution in carbon tetrachloride proceeds via the reaction 2 N2O5(soln) → 4 NO2(soln) + O2(soln) Th
Fantom [35]

<u>Answer:</u> The amount remained after 151 seconds are 0.041 moles

<u>Explanation:</u>

All the radioactive reactions follows first order kinetics.

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,  

k = rate constant  = 4.82\times 10^{-3}s^{-1}

t = time taken for decay process = 151 sec

[A_o] = initial amount of the reactant = 0.085 moles

[A] = amount left after decay process =  ?

Putting values in above equation, we get:

4.82\times 10^{-3}=\frac{2.303}{151}\log\frac{0.085}{[A]}

[A]=0.041moles

Hence, the amount remained after 151 seconds are 0.041 moles

7 0
3 years ago
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