Answer:
The answer to your question is 47.44 g of Oxygen
Explanation:
Data
mass of Ammonia = 14.4 g
mass of Oxygen = ?
Balanced chemical reaction
4NH₃ + 7O₂ ⇒ 4NO₂ + 6H₂O
Process
1.- Calculate the molar mass of Ammonia
NH₃ = 4[(1 x 14) + (3 x 1)] = 4[14 + 3] = 4[17] = 68 g
2.- Calculate the molar mass of Oxygen
O₂ = 7[16 x 2] = 7[32] = 224 g
3.- Use proportions to calculate the mass of Oxygen
68g of NH₃ --------------------- 224 g of O₂
14.4 g of NH₃ ----------------- x
x = (14.4 x 224) / 68
x = 3225.6/ 68
x = 47.44 g
Answer:
-177.9 kJ.
Explanation:
Use Hess's law. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2Ca(s) + O2(g) → 2CaO(s) ΔH = -1269.8 kJ We need to get rid of the Ca and O2 in the equations, so we need to change the equations so that they're on both sides so they "cancel" out, similar to a system of equations. I changed the second equation. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ The sign changes in the second equation above since the reaction changed direction. Next, we need to multiply the first equation by two in order to get the coefficients of the Ca and O2 to match those in the second equation. We also multiply the enthalpy of the first equation by 2. 2Ca(s) + 2CO2(g) + O2(g) → 2CaCO3(s) ΔH = -1625.6 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ Now we add the two equations. The O2 and 2Ca "cancel" since they're on opposite sides of the arrow. Think of it more mathematically. We add the two enthalpies and get 2CaO(s) + 2CO2(g) → 2CaCO3(s) and ΔH = -355.8 kJ. Finally divide by two to get the given equation: CaO(s) + CO2(g) → CaCO3(s) and ΔH = -177.9 kJ.
The grams of glucose are needed to prepare 400g of a 2.00%(m/m) glucose solution g is calculated as below
=% m/m =mass of the solute/mass of the solution x100
let mass of solute be represented by y
mass of solution = 400 g
% (m/m) = 2% = 2/100
grams of glucose is therefore =2/100 = y/400
by cross multiplication
100y = 800
divide both side by 100
y= 8.0 grams
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Answer:
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Explanation:
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