As,
5471 kJ heat is given by = 1 mole of Octane
Then,
5310 kJ heat will be given by = X moles of Octane
Solving for X,
X = (5310 kJ × 1 mol) ÷ 5471 kJ
X = 0.970 moles of Ocatne
So, 0.970 moles of Octane will liberate 5310 kJ energy. Now changing moles to mass,
As,
Moles = mass / M.mass
Or,
Mass = Moles × M.mass
Putting values,
Mass = 0.970 mol × 114.23 g/mol
Mass = 110.83 g of Octane
Answer is: C₃H₃N₃O₃.
Chemical reaction: CₓHₓNₓOₓ + O₂ → aCO₂ + x/2H₂ + x/2N₂.
m(CₐHₓNₓ) = 5,214 g.
m(CO₂) = 5,34 g.
m(H₂) = 1,09 g.
m(N₂) = 1,70 g.
n(CO₂) = n(C) = 5,34 g ÷ 44 g/mol = 0,121 mol.
n(H₂O) = 1,09 g ÷18 g/mol = 0,06 mol.
n(H) = 2 · 0,0605 mol = 0,121 mol.
n(N₂) = 1,7 g ÷ 28 g/mol = 0,0607 mol.
n(N) = 0,0607 mol · 2 = 0,121 mol.
n(C) : n(H) : n(N) = 0,121 mol : 0,121 mol : 0,121 mol /: 0,121
n(C) : n(H) : n(N) = 1 : 1 : 1.
M(CHN) = 27 g/mol.
m(O₂) = 8,13 g - 5,214 g = 2,914 g.
n(O₂) = 2,914 g ÷ 32 g/mol = 0,09 mol.
n(CₓHₓNₓOₓ) = 5,214 g ÷ 129,1 g/mol = 0,0404 mol.
n(CₓHₓNₓOₓ) : n(CO₂) = 1 : 3.
Answer:
(a) 0.047 g (b) 0.0016 oz (c) 0.0001 lb
Explanation:
The given mass of the sodium in the slice = 47 mg
(a) Mass has to be calculated in grams
The conversion of mg to g is shown below as:
1 mg = 10⁻³ g
So,
<u>Mass of sodium = 47 × 10⁻³ g = 0.047 g</u>
(b) Mass has to be calculated in ounces
The conversion of ounces to g is shown below as:
453.6 g = 16 oz
Or,
1 g = 16 / 453.6 oz
So,
<u>Mass of sodium = (0.047 × 16) / 453.6 oz = 0.0016 oz</u>
(c) Mass has to be calculated in pounds
The conversion of pounds to g is shown below as:
1 lb = 453.6 g
Or,
1 g = 1/ 453.6 lb
So,
<u>Mass of sodium = (0.047 × 1) / 453.6 oz = 0.0001 lb</u>
To solve for the number of moles, we simply have to use the Avogadros number which states that there are 6.022 x 10^23 molecules per mole. Therefore:
number of moles = 6.67 X 10^40 chlorine molecules / (6.022 x 10^23 molecules / mole)
number of moles = 1.108 x 10^17 moles
Molarity = moles of solute/volume of solution in liters.
From this relation, we can figure out the number of moles of solute by multiplying the molarity of the solution by the volume in liters.
We have 53.1 mL, or 0.0531 L, of a 12.5 M, or 12.5 mol/L, solution. Multiplying 12.5 mol/L by 0.0531 L, we obtain 0.664 moles. So, in this volume of solution, there are 0.664 moles of solute (HCl).
