Answer:
a)23.2 L
b)68.3kPa
c)7.5 atm
d)60.5L
e)1.67 atm
Explanation:
From Boyle's law:
P1V1=P2V2
P1= 748mmHg
P2=725mmHg
V1= 22.5L
V2??
V2= P1V1/P2= 748×22.5/725= 23.2 L
b)
V1=4.0L
P1= 205×10^3Pa
V2= 12.0L
P2=???
P2= P1V1/V2= 205×10^3×4/12
P2= 68.3×10^3 Pa or 68.3kPa
c)
P1= 1 atm
V1= 196.0L
P2= ??
V2= 26.0L
P2= P1V1/V2=1×196.0/26.0
P2= 7.5 atm
d)
V1= 40.0L
P1= 12.7×10^3Pa
V2=???
P2= 8.4×103Pa
V2= P1V1/P2= 12.7×10^3×40.0/8.4×103
V2=60.5L
e)
V1= 100mL
P1= 1atm
V2= 60mL
P2=???
P2= P1V1/V2= 1×100/60
P2= 1.67 atm
Answer:
The answer to your question is: 24 grams of D
Explanation:
To answer this question we need to remember the Lavoisier law of conservation of mass, which says that in a chemical reaction matter is neither created nor destroyed.
This means that the amount of matter stays the same.
Then, the reaction is
A + B ⇒ C + D
26 g 12 g 14 g x
mass
of reactants 38 g ? mass of products, but it must be
equal to the mass of products
Then 14g + x = 38
x = 38 - 14
x = 24 g of D
Given:
0.607 mol of the weak acid
0.609 naa
2.00 liters of solution
The solution for finding the ph of a buffer:
[HA] = 0.607 / 2.00 = 0.3035 M
[A-]= 0.609/ 2.00 = 0.3045 M
pKa = 6.25
pH = 6.25 + log 0.3045/ 0.3035 = 6.25 is the ph buffer prepared.
To change only one variable which is very important than to test the experiment to match the hypothesis again, I think. It’s been a while since I was on that lesson♀️
chlorine
