Consider the equation -- 2NaI(aq) + Cl_2(g) rightarrow I_2 (aq) + 2NaCl(aq). The
ID: 1009572 • Letter: C
Question
Consider the equation -- 2NaI(aq) + Cl_2(g) rightarrow I_2 (aq) + 2NaCl(aq). The special undergoing reduction is sodium. iodide. chlorine. iodine. water. The reducing agent (if any) in the following equation is: 2Mg (s) + TICl_4(l) rightarrow Ti(s) + 2MgCl_2(s) Mg(s) TiCl_4 (t) Ti(s) MgCl_2(s) not a redox reaction The oxidation number of chlorine in KClO_3 is +6. +5. -1. -2. +2. Which of the following equations represents an oxidation-reduction reaction 3Mg(s) + N_2(g) rightarrow Mg_3 N_2(s) H2_CO_3(aq) rightarrow H_2O(t) + CO_2(g) Sr (NO_3)_2 (aq) + Na_2SO_4(aq) rightarrow SrSO_4 (s) + 2NaNO_3(aq) 1 2 and 3 1 and 3 1 and 2 1, 2, and 3 Which of the following produces radiation of the highest frequency? x-rays AM radio FM radio microwave oven radarExplanation / Answer
Que 1)
answer a) sodium
chlorine and iodine are zero. pure elements that arent bonded have a 0 redox number.
sodium is 2, the Iodine in NaI is -2, sodium in sodium chloride is 2 and chlorine is 2-
chlorine is reduced, and iodine is oxidized
Que 2)
answer a) Mg (s)
Magnesium changes to Mg2+ , Ti4+ changes to Ti. So, Mg is oxidized to Mg +2 and Ti4+ is reduced to Ti
Half-reactions:
Mg(l) <-> Mg2+(l) + 2 e-…..x4..(OX)
Ti+4(g) + 4 e- <-> Ti(s)…..x2..(RED)
Que 3)
answer : b) +5
First, let's look at the K (potassium). In an ionic compound, the potassium's oxidation state is always +1.
Next, let's look at the chlorate , ClO3. The charge on the polyatomic ion is 1. You may be asking yourself, how do I know its charge will be 1?
Since the entire compound KClO3 has a charge of 0 and the K will have a charge of +1, the ClO3 must balance the K's +1 in the form of 1 for a nat charge of 0.
When oxygen is with another element that is less electronegative than it is, the charge on the oxygen is 2. There are 3 oxygen atoms in the chlorate ion, for a total of 6 charge on the total of the 3 oxygen atoms.
Thus, charge of Cl +(6)=1. That means that the charge on chlorine in potassium chlorate is +5.
another way to calculate
Oxygen has an oxidation number of -2
Since there are 3 oxygens in the reactant the oxidation number becomes -6
Potassium (K) has an oxidation number of +1
Add the oxidation numbers together:
KClO3
(+1) + (x) +(-2 * 3) = 0
x = (+6-1)
x = +5
There is no charge on this molecule so the oxidation numbers of all elements added together must be 0
So far the oxidation number is -5 so the only number that chlorine can be to make the molecule 0 is +5
So the oxidation number of chlorine in the reactant is +5
Que 4)
answer a) 1
explaination
3Mg(s) + N2(g) --------> Mg3N2 (s)
type of reaction : redox reaction
half reactions
2 Mg(s) ------> 3 Mg 2+ + 6 e- oxidation half reaction
N2(g) + 6 e- ----------> N2 -3 reduction half reaction
H2CO3(aq) ----> CO2(g) + H2O(l)
decomposition reaction
Balanced equation:
Na2So4 + Sr(No3)2 = 2 NaNo3 + SrSo4
Reaction type: double replacement
Que 5)
answer : a) X-ray
as per frequency X-ray > Radar > microwave > FM radio >Am radio