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You put in a variable to substitute the unknown number.
Answer:
a. 4.41 g of Urea
b. 1.5 g of Urea
Explanation:
To start the problem, we define the reaction:
2NH₃ (g) + CO₂ (g) → CH₄N₂O (s) + H₂O(l)
We only have mass of ammonia, so we assume the carbon dioxide is in excess and ammonia is the limiting reactant:
2.6 g . 1mol / 17g = 0.153 moles of ammonia
Ratio is 2:1. 2 moles of ammonia can produce 1 mol of urea
0.153 moles ammonia may produce, the half of moles
0153 /2 = 0.076 moles of urea
To state the theoretical yield we convert moles to mass:
0.076 mol . 58 g/mol = 4.41 g
That's the 100 % yield reaction
If the percent yield, was 34%:
4.41 g . 0.34 = 1.50 g of urea were produced.
Formula is (Yield produced / Theoretical yield) . 100 → Percent yield
Answer:
Mass of Sodium = 574.75 g
Mass of Chlorine = 886.25 g
Explanation:
The balance chemical equation for the synthesis of NaCl is,
2 Na + Cl₂ → 2 NaCl
Step 1: <u>Find out moles of each reactant required,</u>
According to balance chemical equation,
2 moles of NaCl is produced by = 2 moles of Na
So,
25 moles of NaCl will be produced by = X moles of Na
Solving for X,
X = 25 mol × 2 mol / 2 mol
X = 25 moles of Na
Similarly for Cl₂,
According to balance chemical equation,
2 moles of NaCl is produced by = 1 mole of Cl₂
So,
25 moles of NaCl will be produced by = X moles of Cl₂
Solving for X,
X = 25 mol × 1 mol / 2 mol
X = 12.5 moles of Cl₂
Step 2: <u>Convert each moles to mass as;</u>
Mass = Moles × Atomic Mass
For Na,
Mass = 25 mol × 22.99 g/mol
Mass = 574.75 g
For Cl₂,
Mass = 12.5 mol × 70.90 g/mol
Mass = 886.25 g