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:
0.4694 moles of CrCl₃
Explanation:
The balanced equation is:
Cr₂O₃(s) + 3CCl₄(l) → 2CrCl₃(s) + 3COCl₂(aq)
The stoichiometry of the equation is how much moles of the substances must react to form the products, and it's represented by the coefficients of the balanced equation. So, 1 mol of Cr₂O₃ must react with 3 moles of CCl₄ to form 2 moles of CrCl₃ and 3 moles of COCl₂.
The stoichiometry calculus must be on a moles basis. The compounds of interest are Cr₂O₃ and CrCl₃. The molar masses of the elements are:
MCr = 52 g/mol
MCl = 35.5 g/mol
MO = 16 g/mol
So, the molar mass of the Cr₂O₃ is = 2x52 + 3x35.5 = 210.5 g/mol.
The number of moles is the mass divided by the molar mass, so:
n = 49.4/210.5 = 0.2347 mol of Cr₂O₃.
For the stoichiometry:
1 mol of Cr₂O₃ ------------------- 2 moles of CrCl₃
0.2347 mol of Cr₂O₃----------- x
By a simple direct three rule:
x = 0.4694 moles of CrCl₃
The volume of base required to completely neutralize the acid is 3.2 mL of NaOH.
The equation of the reaction is;
2NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2H2O(l)
From the question;
Concentration of acid CA = 0.426M
Concentration of base CB = 2.658M
Volume of acid VA = 10.00mL
Volume of base VB = ?
Number of moles of acid NA = 1
Number of moles of base NB = 2
Using the relation;
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
VB = CAVANB/CBNA
VB = 0.426M × 10.00mL × 2/ 2.658M × 1
VB = 3.2 mL
Learn more: brainly.com/question/6111443
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Hibiscus I
Hi yes
The answer is [Ne] 3s^2 3p^5 because chlorine is the fifth element in the 3rd row of elements in in p orbital
