48.3 g AgNO3 / 169.9 g/mol = 0.284 moles AgNO3
0.284 mol AgNO3 X (1 mol Ag2CrO4/2 mol AgNO3) = 0.142 mol Ag2CrO4
0.142 mol Ag2CrO4 X 331.7 g/mol = 47.1 g Ag2CrO4
The oxidation number of elements in equation below are,
4NH₃ + 3Ca(ClO)₂ → 2N₂ + 6H₂O + 3CaCl₂
O.N of N in NH₃ = -3
O.N of Ca in Ca(ClO)₂ and CaCl₂ = +2
O.N of N in N₂ = 0
O.N of Cl in Ca(ClO)₂ = +1
O.N of Cl in CaCl₂ = -1
Oxidation:
Oxidation number of Nitrogen is increasing from -3 (NH₃) to 0 (N₂).
Reduction:
Oxidation number of Cl is decreasing from +1 [Ca(ClO)₂] to -1 (CaCl₂).
Result:
<span>N is oxidized and Cl is reduced.</span>
Answer:
MgCl₂+ Na₂CO₃ ==> MgCO₃ + NaCl
From a quick observation
You see that the right hand side of the eqn is deficient of Sodium and Chlorine
Simply Add a Coefficient of 2 to NaCl to balance it with the left.
Your answer now becomes
MgCl₂ + Na₂CO₃ ==> MgCO₃ + 2NaCl.✅
To answer this question a balanced chemical equation is necessary. The correct equation is: N2 + 3H2 = 2NH3
From this equation, one mole of nitrogen react with 3 moles of hydrogen to give 2 moles of ammonia.
Therefore, the mole ratio of NH3 to N2 is 2:1
Answer:
36365.4 Joules
Explanation:
The quantity of Heat Energy (Q) released on cooling a heated substance depends on its Mass (M), specific heat capacity (C), and change in temperature (Φ)
Thus, Q = MCΦ
Since, M = 45.4 g
C = 3.56 J/g°C,
Φ = 250°C - 25°C = 225°C
Q = 45.4g x 3.56J/g°C x 225°C
Q= 36365.4 Joules
Thus, 36365.4 Joules of heat energy is released when the lithium is cooled.
