Answer:
B) Iron (c=0.45 J/g°C)
Explanation:
Given that:-
Heat gain by water = Heat lost by metal
Thus,
Where, negative sign signifies heat loss
Or,
For water:
Mass = 120 g
Initial temperature = 21.8 °C
Final temperature = 24.5 °C
Specific heat of water = 4.184 J/g°C
For metal:
Mass = 40.2 g
Initial temperature = 99.3 °C
Final temperature = 24.5 °C
Specific heat of metal = ?
So,
<u>This value corresponds to iron. Thus answer is B.</u>
Answer:
10 molecules of NH₃.
Explanation:
N₂ + 3H₂ --> 2NH₃
As the N₂ supply is unlimited, what we need to do to solve this problem is <u>convert molecules of H₂ into molecules of NH₃</u>. To do so we use the <em>stoichiometric coefficients</em> of the balanced reaction:
- 15 molecules H₂ *
= 10 molecules NH₃
10 NH₃ molecules could be prepared from 15 molecules of H₂ and unlimited N₂.
Answer:
The correct answer is CaO > LiBr > KI.
Explanation:
Lattice energy is directly proportional to the charge and is inversely proportional to the size. The compound LiBr comprises Li+ and Br- ions, KI comprises K+ and I- ions, and CaO comprise Ca²⁺ and O²⁻ ions.
With the increase in the charge, there will be an increase in lattice energy. In the given case, the lattice energy of CaO will be the highest due to the presence of +2 and -2 ions. K⁺ ions are larger than Li⁺ ion, and I⁻ ions are larger than Br⁻ ion.
The distance between Li⁺ and Br⁻ ions in LiBr is less in comparison to the distance between K⁺ and I⁻ ions in KI. As a consequence, the lattice energy of LiBr is greater than KI. Therefore, CaO exhibits the largest lattice energy, while KI the smallest.
Answer:
C
Explanation:
The oxidation number of Sulphur in SO4^2- is;
x + 4(-2) = -2
x - 8 = -2
x = -2 + 8
x = 6
Now,
the oxidation number of sulphur in H2SO3 is
2 (1) + x + 3(-2) = 0
2 + x -6 = 0
-4 + x = 0
x = 4
Hence, the oxidation number of sulphur changed from +6 to +4 which signifies gain of two electrons as shown in option C.
A Beta particles is emitted when an atom of 85Kr spontaneously decays.
