Answer:
Molecular formula = C20H30
Explanation:
NB 440mg = 0.44g, 135mg= 0.135g
From the question, moles of CO2= 0.44/44= 0.01mol
Since 1 mol of CO2 contains 1mol of C, it implies mol of C = 0.01
Also from the question, moles of H2O = 0.135/18= 0.0075mole
Since 1 mol of H2O contains 2mol of H, it implies mol of H = 0.0075×2= 0.015 mol of H
To get the empirical formula, divide by smallest number of mole
Mol of C = 0.01/0.01=1
Mol of H = 0.015/0.01= 1.5
Multiply both by 2 to obtain a whole number
Mol of C =1×2 = 2
Mol of H= 1.5×2 = 3
Empirical formula= C2H3
[C2H3] not = 270
[ (2×12) + 3]n = 270
27n = 270
n=10
Molecular formula= [C2H3]10= C20H30
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Explanation:
Compressors and limiters are used to reduce dynamic range — the span between the softest and loudest sounds. Using compression can make your tracks sound more polished by controlling maximum levels and maintaining higher average loudness.
Using the ideal gas law equation, we can find the number of H₂ moles produced.
PV = nRT
Where P - pressure - 0.811 atm x 101 325 Pa/atm = 82 175 Pa
V - volume - 58.0 x 10⁻³ m³
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 32 °C + 273 = 305 K
substituting these values in the equation,
82 175 Pa x 58.0 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 305 K
n = 1.88 mol
The balanced equation for the reaction is as follows;
CaH₂(s) + 2H₂O(l) --> Ca(OH)₂(aq) + 2H₂(g)
stoichiometry of CaH₂ to H₂ is 1:2
When 1.88 mol of H₂ is formed , number of CaH₂ moles reacted = 1.88/2 mol
therefore number of CaH₂ moles reacted = 0.94 mol
Mass of CaH₂ reacted - 0.94 mol x 42 g/mol = 39.48 g of CaH₂ are needed
