I believe that the answer is D.
I hope this helps. :)
Answer:
just replace the 9 mole with 3.68 g of Al .
I think it will help you.
Assuming that the contents of the chamber ar ideal gases. We can use the relation PV=nRT. At a constant
temperature and number of moles of the gas the product of PV is equal to some
constant. At another set of condition of temperature, the constant is still the
same. Calculations are as follows:
P1V1 =P2V2
P2 = (1)(450)/ 48
P2 = 9.375 atm
Answer:
2.75M
Explanation:
Concentration(M)= mol/volume
0.0688mol/0.250L
=0.2752M
Answer:
656.25 mL
Explanation:
M1 = 0.40 M
M2 = 1.50 M
V2 = 175 mL
Plug those values into the equation:
M1V1 = M2V2
(0.40)V1 = (1.50)(175)
---> V1 = 656.25 mL
