47 Kpa would bw the pressure needes
Leftover: approximately 11.73 g of sulfuric acid.
<h3>Explanation</h3>
Which reactant is <em>in excess</em>?
The theoretical yield of water from Al(OH)₃ is lower than that from H₂SO₄. As a result,
- Al(OH)₃ is the limiting reactant.
- H₂SO₄ is in excess.
How many <em>moles</em> of H₂SO₄ is consumed?
Balanced equation:
2 Al(OH)₃ + 3 H₂SO₄ → Al₂(SO₄)₃ + 6 H₂O
Each mole of Al(OH)₃ corresponds to 3/2 moles of H₂SO4. The formula mass of Al(OH)₃ is 78.003 g/mol. There are 15 / 78.003 = 0.19230 moles of Al(OH)₃ in the five grams of Al(OH)₃ available. Al(OH)₃ is in excess, meaning that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.
How many <em>grams</em> of H₂SO₄ is consumed?
The molar mass of H₂SO₄ is 98.076 g.mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.076 = 28.289 g.
How many <em>grams</em> of H₂SO₄ is in excess?
40 grams of sulfuric acid H₂SO₄ is available. 28.289 grams is consumed. The remaining 40 - 28.289 = 11.711 g is in excess. That's closest to the first option: 11.73 g of sulfuric acid.
Answer:
the empirical (lowest raios) is
C2H4Cl
Explanation:
A compound is known to consist solely of carbon, hydrogen, and chlorine. Through elemental analysis, it was determined that the compound is composed of 24.27% carbon.
What is the empirical formula of this compound?
the compound has ONLY C, H, and Cl
the % Cl = 100% - 24.27% -4.03% = 71.7%
in 100 gm, there are 71.7 gm Cl, 24.27 gm C, and 4.03 gm H
the number of moles are Cl=71.7/70.91 =1.01= ~ 1
C = 24.27/12.0 = 2.02 =~ 2
H = 403/1.01 = 3.97 =~ 4
so the empirical (lowest raios) is
C2H4Cl
<span>Use Root mean square speed formula:
vrms = Sqrt (3RT/M)
T - temperature in K = 273K - 21CÎż = 252K
R- molar gas constant = 8.31J/Kmol
M - molar mass of a molecule
MN2 = 14x2 = 28 g/mol
MO2 = 16 x 2 = 32 g/mol
MO3 = 16 x 3 = 48 g/mol
Having all info now, you can calculate root mean square speed,vrms ,
for every molecule by using the aboveequasion and only changing M value.</span>
