Yes, it is possible to go do because it would be 2 stacks of 6
Answer:
20L is the new volume
Explanation:
In this case, moles and T° from the gas remain constant. This is the formula we must apply, to solve this:
P₁ . V₁ = P₂ . V₂
5 atm . 10 L = P₂ . 2.5L
P₂ = (5 atm . 10 L) / 2.5L →20L
The dissociation equation will be
NH4OH ---> NH4+ + OH-
Initial 0.006 0 0
Change -0.006 X 0.053 +0.006 X 0.053 -0.006 X 0.053
Equlibrium 0.006 -0.006 X 0.053 0.006 X 0.053 0.006 X 0.053
Ka = [NH4+] [ OH-] / [NH4OH] = (0.006 X 0.053)^2 / 0.006 -0.006 X 0.053
Ka = 1.78 X 10^-5
Answer:

Explanation:
Hello!
In this case, since the density is computed by dividing the mass of the substance by its occupied volume (d=m/V), we first need to realize that 0.8206 g/mL is the same to 0.8206 kg/L, which means we first need to compute the volume in L:

Then, solving for the mass in d=m/V, we get m=d*V and therefore the mass of gasoline in that full tank turns out:

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