There are approximately 4 resonance structures that can be drawn for N205 (no N-N bond) (minimal formal charge).
Answer: The derivative of a constant term is always 0. So the acceleration of the body would be zero. Hence, the acceleration of a body moving with uniform velocity will always be zero.
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Answer:
Na.
Explanation:
- The oxidation-reduction reaction contains a reductant and an oxidant (oxidizing agent).
- An oxidizing agent, or oxidant, gains electrons and is reduced in a chemical reaction. Also known as the electron acceptor, the oxidizing agent is normally in one of its higher possible oxidation states because it will gain electrons and be reduced.
- A reducing agent (also called a reductant or reducer) is an element (such as calcium) or compound that loses (or "donates") an electron to another chemical species in a redox chemical reaction.
<em>2Na + S → Na₂S.</em>
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Na is oxidized to Na⁺ in (Na₂S) (loses 1 electron). "reducing agent".
S is reduced to S²⁻ in (Na₂S) (gains 2 electrons). "oxidizing agent".
Concentration of the reactant,pressure,surface
area of the reactant and temperatur
Given:
Ma = 31.1 g, the mass of gold
Ta = 69.3 °C, the initial temperature of gold
Mw = 64.2 g, the mass of water
Tw = 27.8 °C, the initial temperature of water
Because the container is insulated, no heat is lost to the surroundings.
Let T °C be the final temperature.
From tables, obtain
Ca = 0.129 J/(g-°C), the specific heat of gold
Cw = 4.18 J/(g-°C), the specific heat of water
At equilibrium, heat lost by the gold - heat gained by the water.
Heat lost by the gold is
Qa = Ma*Ca*(T - Ta)
= (31.1 g)*(0.129 J/(g-°C)(*(69.3 - T °C)-
= 4.0119(69.3 - T) j
Heat gained by the water is
Qw = Mw*Cw*(T-Tw)
= (64.2 g)*(4.18 J/(g-°C))*(T - 27.8 °C)
= 268.356(T - 27.8)
Equate Qa and Qw.
268.356(T - 27.8) = 4.0119(69.3 - T)
272.3679T = 7738.32
T = 28.41 °C
Answer: 28.4 °C
