Answer:
1. Cu
2. Cu
3. 2 electrons.
Explanation:
Step 1:
The equation for the reaction is given below:
3Cu(s) + 8HNO3(aq) -> 2NO(g) 3Cu(NO3)2(aq) + 4H2O(l)
Step 2:
Determination of the change of oxidation number of each element present.
For Cu:
Cu = 0 (ground state)
Cu(NO3)2 = 0
Cu + 2( N + 3O) = 0
Cu + 2(5 + (3 x -2)) =0
Cu + 2 (5 - 6) = 0
Cu + 2(-1) = 0
Cu - 2 = 0
Cu = 2
The oxidation number of Cu changed from 0 to +2
For N:
HNO3 = 0
H + N + 3O = 0
1 + N + (3 x - 2) = 0
1 + N - 6 = 0
N = 6 - 1
N = 5
NO = 0
N - 2 = 0
N = 2
The oxidation number of N changed from +5 to +2
The oxidation number of oxygen and hydrogen remains the same.
Note:
1. The oxidation number of Hydrogen is always +1 except in hydride where it is - 1
2. The oxidation number of oxygen is always - 2 except in peroxide where it is - 1
Step 3:
Answers to the questions given above
From the above illustration,
1. Cu is oxidize because its oxidation number increased from 0 to +2 as it loses electron.
2. Cu is the reducing agent because it reduces the oxidation number of N from +5 to +2.
3. The reducing agent i.e Cu transferred 2 electrons to the oxidising agent HNO3 because its oxidation number increase from 0 to +2 as it loses its electrons. This means that Cu transfer 2 electrons.
Answer:
MCl₂
Explanation:
The formula for boiling point elevation can be used to find x. The "complete dissociation" means there will be an ion of M and x ions of Cl in the solution. The number of moles of solute will be 30.2 grams divided by the molecular weight of MClx, where x is the variable we're trying to find.
Then the formula for the salt is MCl₂.
Answer:
B,C,D
Explanation:
The yield of CCl4 depends on the amount of CH4 in a 1:1 ratio. The amount of Cl2 is twice that of CH4 hence some must be left over. To ensure that all the Cl2 is used up, more CH4 must added to the system.
Answer:
3 acidic hydrogens per molecule of citric acid
Explanation:
In a sample of 37.2 mL(0.0372 L) of 0.105 mol/L of NaOH, will have:
n = 0.105x0.0372 = 0.0039 mol of NaOH
The dissociation of NaOH will give the same number of moles of Na⁺ and OH⁻.
The molar mass of citric acid is:
C: 12g/mol x 6 = 72 g/mol
H: 1g/mol x 8 = 8g/mol
O: 16 g/mol x 7 = 112 g/mol
192 g/mol
So, 0.250g of the acid has
n = mass/molar mass
n = 0.250/192
n = 0.0013 mol.
To be neutralized, it will be necessary 0.0039 mol of acidic hydrogens to react with the 0.0039 mol of OH⁻.
The dissociation reaction of one molecule of the acid will give the stoichiometry:
1 mol of acid ----------------------- x mol of acidic hydrogens
0.0013 mol --------------------------- 0.0039
For a simple direct three rule:
0.0013x = 0.0039
x = 3 acidic hydrogens per molecule of citric acid.
NH4I (aq) + KOH (aq) in chemical equation gives
NH4I (aq) + KOH (aq) = KI (aq) + H2O(l) + NH3 (l)
Ki is in aqueous state H2o is in liquid state while NH3 is in liquid state
from the equation above 1 mole of NH4I (aq) react with 1 mole of KOH(aq) to form 1mole of KI(aq) , 1mole of H2O(l) and 1 Mole of NH3(l)
