Oxidation number of an atom is the charge that atom would have if the compound is composed of ions. In neutral substances that contains atoms of one element the oxidation number of an atom is zero. Thus atoms in O2, Ni2, and aluminium all have oxidation number of zero.
In this case, Ni2, the oxidation number of Ni atom is zero,
for NiO4-, assuming oxidation number of Ni is x
(x ×1) + (-2 × 4) = -1
x = + 7
Therefore, the oxidation number goes from 0 to +7
Answer:
Please find the complete question and its solution in the attached file:
Explanation:
Answer:
296.1 day.
Explanation:
- The decay of radioactive elements obeys first-order kinetics.
- For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).
Where, k is the rate constant of the reaction.
t1/2 is the half-life time of the reaction (t1/2 = 1620 years).
∴ k = ln2/(t1/2) = 0.693/(74.0 days) = 9.365 x 10⁻³ day⁻¹.
- For first-order reaction: <em>kt = lna/(a-x).</em>
where, k is the rate constant of the reaction (k = 9.365 x 10⁻³ day⁻¹).
t is the time of the reaction (t = ??? day).
a is the initial concentration of Ir-192 (a = 560.0 dpm).
(a-x) is the remaining concentration of Ir-192 (a -x = 35.0 dpm).
<em>∴ kt = lna/(a-x)</em>
(9.365 x 10⁻³ day⁻¹)(t) = ln(560.0 dpm)/(35.0 dpm).
(9.365 x 10⁻³ day⁻¹)(t) = 2.773.
<em>∴ t </em>= (2.773)/(9.365 x 10⁻³ day⁻¹) =<em> 296.1 day.</em>
Explanation:
It is known that efficiency is denoted by
.
The given data is as follows.
= 0.82,
= (21 + 273) K = 294 K
= 200 kPa,
= 1000 kPa
Therefore, calculate the final temperature as follows.
0.82 =
= 1633 K
Final temperature in degree celsius =
= 
Now, we will calculate the entropy as follows.

For 1 mole, 
It is known that for
the value of
= 0.028 kJ/mol.
Therefore, putting the given values into the above formula as follows.

= 
= 0.0346 kJ/mol
or, = 34.6 J/mol (as 1 kJ = 1000 J)
Therefore, entropy change of ammonia is 34.6 J/mol.
thanks for the answers ッ. (btw they’re on the bottom of the question if anyone doesn’t see it.