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alukav5142 [94]
3 years ago
10

Under what pressure will a scientist need to store 0.400 moles of gas if the container has a volume of 200.0 mL and the temperat

ure is kept at 23.0 C?
Chemistry
2 answers:
STALIN [3.7K]3 years ago
5 0

Answer: 48.60 atm

Explanation: Using ideal gas equation:

PV=nRT

P=\frac{nRT}{V}

P= pressure = ? atm

V= volume =200ml=0.2L  

n = no of moles = 0.400

R= gas constant =0.0821 Latm/molK

T = Temperature =23^0C=(23+273)K=296K

P=\frac{0.400\times 0.0821\times 296K}{0.2L}=48.60atm

Nutka1998 [239]3 years ago
4 0
P*V=n*R*T where p is the pressure, v the volume in L, n the quantity in moles, R a standard number that should be given and T the temperature in Kelvin
R=8.32 (usually)
T(in K)=273+T(in Celsius) so T=296K
P*V=n*R*T
P=n*R*T/V
P=0.4*8.32*296/0.2
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4 years ago
What volume (in milliliters) of oxygen gas is required to react with 4.03 g of Mg at STP?
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The first step is to find the number of moles of Mg in 4.03g of Mg.  You can do this by dividing 4.03g Mg by its molar mass (which is 24.3g/mol) to get 0.1658mol Mg.  Then you have to find the number of moles of O₂ that will react with 0.1658mol Mg.  To do this you need to use the fact that 1mol O₂ will react with 2mol Mg (this reatio is from the chemical equation) so you have to multiply 0.1658mol Mg by (1mol O₂)/(2mol Mg) to get 0.0829mol O₂.  From here you would usually use PV=nRT and solve for V However, the question tells us that we are at STP, that means you can use the fact that 22.4L of gas is 1 mol of gas at STP.  Using that information we can find the volume of O₂ gas by mulitlying 0.0829mol O₂ by 22.4L/mol to get 1.857L which equals 1857mL.
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I hope this helps. Let me know in the comments if anything is unclear.
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