Answer:
65.08 g.
Explanation:
- For the reaction, the balanced equation is:
<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>
2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.
- Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:
<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>
<u><em>Using cross multiplication:</em></u>
2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.
0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.
∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.
<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol
) = <em>65.08 g.</em>
Answer:
The answer to your question is 8.74 g of He
Explanation:
Data
V = 2.4 x 10² L
P = 99 kPa
T = 0°C
mass = ?
Process
1.- Convert kPa to atm
P = 99 kPa = 99000 Pa
1 atm --------------- 101325 Pa
x --------------- 99000 Pa
x = (99000 x 1) / 101325
x = 0.977 atm
2.- Convert temperature to °K
°K = 273 + 0
°K = 273
3.- Substitution
PV = nRT
- Solve for n
n = PV / RT
n = (0.977)(2.4 x 10²) / (0.082)(273)
n = 24.48 / 22.386
n = 1.093 moles
4.- Calculate the grams of He
8 g -------------------- 1 mol
x -------------------- 1.093 moles
x = (1.093 x 8) / 1
x = 8.74 g
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