Umm its not a i think... lemme see lol
Answer: Volume would be 196.15 mL if the temperature were changed to 
and the pressure to 1.25 atmospheres.
Explanation:
Given: 
, 
= 256 mL,
= 720 torr (1 torr = 0.00131579 atm) = 0.947368 atm
, 
Formula used to calculate volume is as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that the volume would be 196.15 mL if the temperature were changed to and the pressure to 1.25 atmospheres.
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>
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Answer is: the molar mass od sodium carbonate (Na₂CO₃) is 106.0 g/mol.
M(Na₂CO₃) = 2 · Ar(Na) + Ar(C) + 3 · Ar(O).
M(Na₂CO₃) = 2 · 23 + 12 + 3 · 16 · g/mol.
M(Na₂CO₃) = 46 + 12 + 48 · g/mol.
M(Na₂CO₃) = 106 g/mol; molar mass of sodium carbonate.
Ar is relative atomic mass (the ratio of the average mass of atoms of a chemical element to one unified atomic mass unit) of an element.
