Answer:
1.008moles of iodine
Explanation:
Hello,
This question requires us to calculate the theoretical yield of I₂ or number of moles that reacted.
Percent yield = (actual yield / estimated yield) × 100
Actual yield = 1.2moles
Estimated yield = ?
Percentage yield = 84%
84 / 100 = 1.2 / x
Cross multiply and solve for x
100x = 84 × 1.2
100x = 100.8
x = 100.8/100
x = 1.008moles
1.008 moles of I₂ reacted in excess of H₂ to give 1.2 moles of HI
Answer: It is caused by the tendency of some atoms in molecules to attract electrons more than their accompanying atom.
Answer:
i believe its called A. periods
Assuming the volume of the gas is measured at standard temperature and pressure, Then one mole of the Gas would occupy 22.4 liters.
Therefore, 1 liter is 1/224 moles
one mole of nitrogen 14 is 14
Therefore 1 liter of the nitrogen weighs 1/224×14
0.0625 grams
Answer:
[EtOH] = 2.2M and Wt% EtOH = 10.1% (w/w)
Explanation:
1. Molarity = moles solute / Volume solution in Liters
=> moles solute = mass solute / formula weight of solute = 9.8g/46g·mol⁻¹ = 0.213mol EtOH
=> volume of solution (assuming density of final solution is 1.0g/ml) ...
volume solution = 9.81gEtOH + 87.5gH₂O = 97.31g solution x 1g/ml = 97.31ml = 0.09731 Liter solution
Concentration (Molarity) = moles/Liters = 0.213mol/0.09731L = 2.2M in EtOH
2. Weight Percent EtOH in solution (assuming density of final solution is 1.0g/ml)
From part 1 => [EtOH] = 2.2M in EtOH = 2.2moles EtOH/1.0L soln
= {(2.2mol)(46g/mol)]/1000g soln] x 100% = 10.1% (w/w) in EtOH.
