Answer:
[H2] = 0.012 M
[N2] = 0.019 M
[H2O] = 0.057 M
Explanation:
The strategy here is to account for the species at equilibrium given that the concentration of [NO]=0.062M at equilibrium is known and the quantities initially present and its stoichiometry.
2NO(g) + 2H2(g) ⇒ N2(g) + 2H2O(g)
i mol 0.10 0.050 0.10
c mol -0.038 -0.038 +0019 +0.038
e mol 0.062 0.012 00.019 0.057
Since the volume of the vessel is 1.0 L, the concentrations in molarity are:
[NO] = 0.062 M
[H2] = 0.012 M
[N2] = 0.019 M
[H2O] = 0.057 M
The answer is (2) A bond is formed and energy is released. The left side of equation is I atom and the right side of equation is I2 molecule. So the bond is formed between I atom to form I2 molecule. And forming bond will release energy while breaking bond will absorb energy.
When the charged balloon is brought near the wall, it repels some of the negatively charged electrons in that part of the wall. Therefore, that part of the wall is left repelled.
<u>Explanation</u>:
- Balloons don't stick to walls. However, if you rub the balloon on an appropriate piece of material such as clothing or a wall, electrons are pulled from the other material to the balloon.
- The balloon now as more electrons than normal and therefore has an overall negative charge. Two balloons like this will repel each other.
- The other material now has an overall positive charge. Because opposite charges attract, the balloon will now appear to stick to the other material. If you didn't rub the balloon first, it's charge would be neutral and it wouldn't stick to the wall.
Examination by a test; experiment, as in chemistry, metallurgy, etc.
Answer:
The new temperature will be 2546 K or 2273 °C
Explanation:
Step 1: Data given
The initial temperature = 1000 °C =1273 K
The volume = 20L
The volume increases to 40 L
Step 2: Calculate the new temperature
V1/T1 = V2/T2
⇒with V1 = the initial volume = 20L
⇒with T1 = the initial temperature = 1273 K
⇒with V2 = the increased volume = 40L
⇒with T2 = the new temperature = TO BE DETERMINED
20L/ 1273 K = 40L / T2
T2 = 40L / (20L/1273K)
T2 = 2546 K
The new temperature will be 2546 K
This is 2546-273 = 2273 °C
Since the volume is doubled, the temperature is doubled as well
