1 mol of Carbon = 12 grams.
x mol of Carbon = 55 grams
12*x = 1 * 55
x = 55/12
x = 4.583333 mols of carbon
1 mol of anything is 6.02 * 10^23 atoms
4.58333333 mol = x
1/4.5833333 = 6.02 * 10^23/x
x = 4.58333* 6.02*10^23
x = 2.7591 * 10^23 Carbon atoms
Answer:
ΔH⁰(11.4g NH₄NO₃) = -30.59Kj (4 sig. figs. ~mass of NH₄NO₃(s) given) (exothermic)
Explanation:
3NH₄NO₃(s) + C₁₀H₂₂(l) + 14O₂(g) => 3N₂(g) + 17H₂O(g) + 10CO₂(g)
ΔH⁰(f): 3(-365.6)Kj 1(-301)Kj 14(0)Kj 3(0)Kj 17(-241.8)Kj 10(-393.5)Kj
= -1096.8Kj = -301Kj = 0Kj = 0Kj = -4110.6Kj = -3930.5Kj
ΔHₙ°(rxn) = ∑
(ΔH˚(f)products) - ∑(ΔH˚(f)reactants)
= [3(0)Kj + 17(-241.8)Kj + (-393.5)Kj] - [(-(1096.8)Kj + (-301)Kj + (0)Kj]
= [-(8041.1) - (-1397.8)]Kj
= -6643.3Kj (for 3 moles NH₄NO₃ used in above equation)
∴ Standard Heat of Rxn = -6643.3Kj/3moles = -214.8Kj/mole NH₄NO₃(s)
ΔH°(rxn for 14.11g NH₄NO₃(s)) = (11.4g/80.04g·mol⁻¹)(-214.8Kj/mol) = 30.5937Kj ≅ 30.59Kj (4 sig. figs. ~mass of NH₄NO₃(s) given)
Given:
Stock dose/concentration of 20% Acetylcysteine (200 mg/mL)
150 mg/kg dose of Acetylcysteine
Weight of the dog is 13.2 lb
First we must convert 13.2 lb to kg:
13.2 lb/(2.2kg/lb) = 6 kg
Then we must calculate the dose:
(150 mg/kg)(6kg) = 900 mg
Lastly, we must calculate the dose in liquid form to be administered:
(900 mg)/(200 mg/mL) = 4.5 mL
Therefore, 4.5 mL of 20% Acetylcysteine should be given.
Answer:
0.143L
Explanation:
Molar mass of H2SO4 = 98g/Mol
No of mole = mass/molar mass
No of mole= 49/98 = 0.5 mol
No of mol = concentration × volume
Volume = n/C = 0.5/3.5 = 0.143L
