The reaction between K₂SO₄(aq) and SrI₂(aq) produces KI(aq) and SrSO₄(s) as products.
The reaction is
K₂SO₄(aq) + SrI₂(aq) → KI(aq)+ SrSO₄(s)
To balance the equation both side of the reaction should have same number of atoms in each element.
Right hand side of the reaction has 1 K, 1 I, 1 Sr, 1 S and 4 O atoms while 2 K, 2 I, 1 Sr,1 S and 4 O present in left hand side of the reaction.
Hence, number of I atoms and number of K atoms are not balanced.
To balance the K atoms we should add 2 before KI. Then I atoms will be 2 at the right hand side.
Hence, the balanced reaction equation is
K₂SO₄(aq) + SrI₂(aq) → 2KI(aq)+ SrSO₄(s)
Answer:
The final product of the reaction is (<em>2S,3S</em>)-2-ethoxy-3-methylpentane.
Explanation:
The given reaction undergoes 
mechanism in which the nucleophile attacks the backside and it is substituted by the elimination of bromine.
Due to the backside attack of nucleophile , the inverse in stereo-chemistry is observed.
After the substitution of ethoxy group, the configuration is assigned according to the priority it shows clock wise direction(R) - configuration.
When hydrogen faces the front side , it results shows inverse configuration i.e, S- configuration.
The chemical reaction is as follows.
When oxygen is found is peroxide, it has an oxidation number of -1.
The chemical formula of hydrogen peroxide is H2O2. We know that hydrogen always has +1 oxidation state until it forms metal hydrides. So in H2O2, the oxidation state ofhydrogen is +1.
Now, let oxidation state of oxygen be x. So,
2 * (+1) + 2*x = 0
2 + 2x = 0
2x = -2
x = -2 / 2
x = -1
Hence, the oxidation number of oxygen in peroxides is -1
Answer:
Prefixes for carbon chain length are
1 carbon = meth
2 carbon = eth
3 = prop
4 = but
5 = pent
6 = hex
7 = hept
8 = oct
9 = non
10 = dec
Explanation:
