Enthalpy change during the dissolution process = m c ΔT,
here, m = total mass = 475 + 125 = 600 g
c = <span>specific heat of water = 4.18 J/g °C
</span>ΔT = 7.8 - 24 = -16.2 oc (negative sign indicates that temp. has decreases)
<span>
Therefore, </span>Enthalpy change during the dissolution = 600 x 4.18 X (-16.2)
= -40630 kJ
(Negative sign indicates that process is endothermic in nature i.e. heat is taken by the system)
Thus, <span>enthalpy of dissolving of the ammonium nitrate is -40630 J/g</span>
Answer:
the air will escape from the jar
Explanation:
Due to its low density air in the jar will be displaced and be replaced by water.
Answer is: <span>the molarity of the sulfuric acid is 7.14 M.
</span>ω(H₂SO₄) = 50% ÷ 100% = 0.5.<span>
d(H</span>₂SO₄) = 1.4 g/mL.
V(H₂SO₄) = 100 mL ÷ 1000 mL/L = 0.1 L..
mr(H₂SO₄) = d(H₂SO₄) · V(H₂SO₄).
mr(H₂SO₄) = 1.4 g/mL · 100 mL.
mr(H₂SO₄) = 140 g.
m(H₂SO₄) = ω(H₂SO₄) · mr(H₂SO₄).
m(H₂SO₄) = 0.5 · 140 g.
m(H₂SO₄) = 70 g.
n(H₂SO₄) = m(H₂SO₄) ÷ M(H₂SO₄).
n(H₂SO₄) = 70 g ÷ 98 g/mol.
n(H₂SO₄) = 0.714 mol.
c(H₂SO₄) = n(H₂SO₄) ÷ V(H₂SO₄).
c(H₂SO₄) = 0.714 mol ÷ 0.1 L.
C(H₂SO₄) = 7.14 M.
<u>Answer:</u> The correct answer is Option A.
<u>Explanation:</u>
Oxidation reaction is defined as the reaction in which an atom looses its electrons. Here, oxidation state of the atom increases.
Reduction reaction is defined as the reaction in which an atom gains electrons. Here, the oxidation state of the atom decreases.
For the given chemical reaction:
The half reactions for the above reaction are:
Oxidation half reaction: 
Reduction half reaction: 
From the above reactions, copper is loosing its electrons. Thus, it is getting oxidized.
Silver ion is gaining electrons and thus is getting reduced.
Hence, the correct answer is Option A.
Answer:
Your help from me is a good luck! :)
Explanation:
Lol sorry I don't know the answer and don't want to tell you something wrong. Good luck though. Have a great day!
