Answer:
2Al+ 6HNO3 ---- 3H2 + 2Al(NO3)3
Explanation:
Put coefficient a,b,c, and d for calculation:
a Al + b HNO3 = c H2 + d Al(NO3)3
for Al: a = d
for H: b = 2c
for N: b = 3d
for O: 3b = 9d
Suppose a=1, then d=1, b=3, c=3/2
multiply 2 to make all natural number, a=2, then b=6, c=3, d=2
Answer:
The same holds true for pH values above 7, each of which is ten times more alkaline (another way to say basic) than the next lower whole value. For example, pH 10 is ten times more alkaline than pH 9 and 100 times (10 times 10) more alkaline than pH 8.
Explanation:
Answer:
The element will be 
Explanation:
Given that,
Number of proton = 80
Number of neutron = 81
Number of electron = 79
We know that,
The atomic number is equal to the number of proton.
So, the atomic number is 80.
According to atomic number,
The element will be mercury.
We need to calculate the atomic mass
Using formula of atomic mass

Put the value into the formula


We need to find the element
Using atomic mass and atomic number


So, the element will be

Put the value of A and Z
Hence, The element will be 
Answer:
<h3>1. 10 e⁻</h3>
Oxidation numbers
I₂O₅(s): I (5+); O(2-)
CO(g): C(2+); O(2-)
I₂(s): I(0)
CO₂(g): C(4+); O(2-)
<h3>2. 4 e⁻</h3>
Oxidation numbers
Hg²⁺(aq): Hg(2+)
N₂H₄(aq): N(2-); H(1+)
Hg(l): Hg(0)
N₂(g): N(0)
H⁺(aq): H(1+)
<h3>3. 6 e⁻</h3>
Oxidation numbers
H₂S(aq): H(1+); S(2-)
H⁺(aq): H(1+)
NO₃⁻(aq): N(5+); O(2-)
S(s): S(0)
NO(g): N(2+); O(2-)
H₂O(l): H(1+); O(2-)
Explanation:
In order to state the total number of electrons transferred we have to identify both half-reactions for each redox reaction.
1. I₂O₅(s) + 5 CO(g) → I₂(s) + 5 CO₂(g)
Oxidation: 10 e⁻ + 10 H⁺(aq) + I₂O₅(s) → I₂(s) + 5 H₂O(l)
Reduction: 5 H₂O(l) + 5 CO(g) → 5 CO₂(g) + 10 H⁺(aq) + 10 e⁻
2. 2 Hg²⁺(aq) + N₂H₄(aq) → 2 Hg(l) + N₂(g) + 4 H⁺(aq)
Oxidation: N₂H₄(aq) → N₂(g) + 4 H⁺(aq) + 4 e⁻
Reduction: 2 Hg²⁺(aq) + 4 e⁻ → 2 Hg(l)
3. 3 H₂S(aq) + 2H⁺(aq) + 2 NO₃⁻(aq) → 3 S(s) + 2 NO(g) + 4H₂O(l)
Oxidation: 3 H₂S(aq) → 3 S(s) + 6 H⁺(aq) + 6 e⁻
Reduction: 8 H⁺(aq) + 2 NO₃⁻(aq) + 6 e⁻ → 2 NO(g) + 4 H₂O
Since K3 is greater than 1, the reaction is thermodynamically favorable.
Data;
- FeF2 is in an acidic solution
- K2 > 1 is thermodynamically favorable
- k3 > 1 is not thermodynamically favorable
- K1<1 is not thermodynamically favorable
<h3>Hess Law</h3>
This states that the heat change in a given chemical reaction is the same irrespective of the number of process in which the reaction is effected.
Using Hess Law;

substituting this using Gibbs law,

Since K3 is greater than 1, the reaction is thermodynamically favorable.
Learn more on Hess law here;
brainly.com/question/12895682