There are TWO atoms in one molecule of hydrogen.
Answer:
Air and Water Temperature Increases
An increase in the air temperature will cause water temperatures to increase as well. ... Lower levels of dissolved oxygen due to the inverse relationship that exists between dissolved oxygen and temperature. As the temperature of the water increases, dissolved oxygen levels decrease.
Answer:
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
Explanation:
Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)
In the oxidation half reaction, the oxidation number increases:
Mn changes from 0, in the ground state to Mn²⁺.
The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.
Silver changes from Ag⁺ to Ag.
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
To balance the hole reaction, we need to multiply by 2, the second half reaction:
Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2
2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)
Now we sum, and we can cancel the electrons:
2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻
Answer:
Vf = 1.22 mL
Explanation:
If we assume that the pressure is constant and the number of moles does not change, we can say that the volume of the gas is modified in a directly ratio, to the Absolute Temperature.
Let's convert the values:
91°C + 273 = 364K
0.9°C + 273 = 273.9K
Volume decreases if the temperature is decreases
Volume increases if the T° increases
V₁ / T₁ = V₂ / T₂ → 1.63mL /364K = V₂ / 273.9K
V₂ = (1.63mL /364K) . 273.9K → 1.22 mL
Answer:
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