<span>We use the formula PV = nRT. P = 758 torr = 0.997 atm. V = 3.50 L. T = 35.6 C = 308.15 K. R = 0.0821. Rearranging the equation gives up n = PV/Rt and we get .0138 moles of butane. Mass of 0.0138 moles of butane = .0138 x 58.12 = 8.02g.</span>
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<h3>
Answer: 144 g</h3>
Explanation:
Mass of glucose = moles × molar mass
∴ Mass of glucose = 0.8 mol × 180 g mol⁻¹
= 144 g
∴ the mass of glucose you need to have 0.8 mol of glucose = 144 g
Answer:
0.677 moles
Explanation:
Take the atomic mass of K = 39.1, O =16.0, P = 31.0
no. of moles = mass / molar mass
no. of moles of K3PO4 used = 4.79 / (39.1x3 + 31 + 16x4)
= 0.02256 mol
From the equation, the mole ratio of KOH : K3PO4 = 3 :1,
meaning every 3 moles of KOH used, produces 1 mole of K3PO4.
So, using this ratio, let the no. of moles of KOH required to be y.

y = 0.02256 x3
y = 0.0677 mol
If you don't find exactly 0.677 moles as one of the options, go for the closest one. A very slight error may occur because of taking different significant figures of atomic masses when calculating.
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