<span>To
solve this we assume that the gas is an ideal gas. Then, we can use the ideal
gas equation which is expressed as PV = nRT. At number of moles the value of PV/T is equal to some constant. At another
set of condition of temperature, the constant is still the same. Calculations
are as follows:</span>
P1V1/T1 = P2V2/T2
P2 = P1 (V1) (T2) / (T1) (V2)
P2 = 475 kPa (4 m^3) (277 K) / (290 K) (6.5 m^3)
P2 = 279.20 kPa
Therefore, the changes in the temperature and the volume lead to a change in the pressure of the system which is from 475 kPa to 279.20 kPa. So, there is a decrease in the pressure.
Answer:
Explanation:
pH = pKa + log [ CH₃COO⁻ ] / [CH₃COOH ]
5.36 = 4.86 + log [ CH₃COO⁻ ] / [CH₃COOH ]
log [ CH₃COO⁻ ] / [CH₃COOH ] = .5
[ CH₃COO⁻ ] / [CH₃COOH ] = 3.16
moles of CH₃COOH = .680 x .9 = .612 M
Let x mole of KOH is required
x /( .612 - x ) = 3.16
x = 1.933 - 3.16 x
x = .46488
.46488 moles of KOH will be required
volume required be V
v x 2.62 = .46488
v = .1774 L
= 177.4 mL
177.4 mL of 2.62 M KOH will be required .
Answer:
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Explanation:
Answer:
527.68 mL
Explanation:
We will assume that nitrogen is behaving as ideal gas here.
For ideal gas the gas law is:
Where
P1= initial pressure = 740 torr
V1= initial volume = 500mL
T1= initial temperature = 25⁰C = 298 K
P2= final pressure = 760 torr
V2= final volume = ?
T2= final temperature = 50⁰C = 323 K
Putting values in the gas law
Final volume = 
Lets make x equal the number times you use the 1/5 ratio alloy and y equal the number of times you use the 3/1 ratio alloy. You can make the equation (x+3y)-(5x+y)=0 and (x+3y)+(5x+y)=350
Then you can make the system of equation of:
-4x+2y=0 (this is y=2x)
6x+4y=350
You can make 6x+8x=350 by through substitution and solve of x.
14x=350
x=25 (which means y=50)
that means that the weight of the 1/5 ratio alloy is 150 pounds.
(25+(5x25))=150
the weight of the 3/1 ratio alloy is 200 pounds.
((50x3)+50)=200
I hope this helps.
