Answer:
The 150 g Al will reach a higher temperature.
Explanation:
- The amount of heat added to a substance (Q) can be calculated from the relation:
Q = m.c.ΔT.
where, Q is the amount of heat added,
m is the mass of the substance,
c is the specific heat of the substance,
ΔT is the temperature difference (final T - initial T).
<em>Since, Q and c is constant, ΔT will depend only on the mass of the substance (m).</em>
∵ ΔT is inversely proportional to the mass of the substance.
∴ The piece with the lowest mass (150.0 g) will reach a higher temperature than that of a higher mass (250.0 g).
<em>So, the right choice is</em>: The 150 g Al will reach a higher temperature.
When 2 moles burn 6 moles of O2 is produced !
so for 16 moles , (6/2)×16= 48 moles will be produced !
so answer is A , 48 moles !
Answer:
0.36 M
Explanation:
There is some info missing. I think this is the complete question.
<em>Suppose a 250 mL flask is filled with 0.30 mol of N₂ and 0.70 mol of NO. The following reaction becomes possible:
</em>
<em>N₂(g) +O₂(g) ⇄ 2 NO(g)
</em>
<em>The equilibrium constant K for this reaction is 7.70 at the temperature of the flask. Calculate the equilibrium molarity of O₂. Round your answer to two decimal places.</em>
<em />
Initially, there is no O₂, so the reaction can only proceed to the left to attain equilibrium. The initial concentrations of the other substances are:
[N₂] = 0.30 mol / 0.250 L = 1.2 M
[NO] = 0.70 mol / 0.250 L = 2.8 M
We can find the concentrations at equilibrium using an ICE Chart. We recognize 3 stages (Initial, Change, and Equilibrium) and complete each row with the concentration or change in the concentration.
N₂(g) +O₂(g) ⇄ 2 NO(g)
I 1.2 0 2.8
C +x +x -2x
E 1.2+x x 2.8 - 2x
The equilibrium constant (K) is:
![K=7.70=\frac{[NO]^{2}}{[N_{2}][O_{2}]} =\frac{(2.8-2x)^{2} }{(1.2+x).x}](https://tex.z-dn.net/?f=K%3D7.70%3D%5Cfrac%7B%5BNO%5D%5E%7B2%7D%7D%7B%5BN_%7B2%7D%5D%5BO_%7B2%7D%5D%7D%20%3D%5Cfrac%7B%282.8-2x%29%5E%7B2%7D%20%7D%7B%281.2%2Bx%29.x%7D)
Solving for x, the positive one is x = 0.3601 M
[O₂] = 0.3601 M ≈ 0.36 M
<u>Given:</u>
Concentration of Cr2+ = 0.892 M
Concentration of Fe2+ = 0.0150 M
<u>To determine:</u>
The cell potential, Ecell
<u>Explanation:</u>
The half cell reactions for the given cell are:
Anode: Oxidation
Cr(s) ↔ Cr2+(aq) + 2e⁻ E⁰ = -0.91 V
Cathode: Reduction
Fe2+ (aq) + 2e⁻ ↔ Fe (s) E⁰ = -0.44 V
------------------------------------------
Net reaction: Cr(s) + Fe2+(aq) ↔ Cr2+(aq) + Fe(s)
E°cell = E°cathode - E°anode = -0.44 - (-0.91) = 0.47 V
The cell potential can be deduced from the Nernst equation as follows:
Ecell = E°cell - (0.0591/n)log[Cr2+]/[Fe2+]
Here, n = number of electrons = 2
Ecell = 0.47 - 0.0591/2 * log[0.892]/[0.0150] = 0.418 V
Ans: The cell potential is 0.418 V
