Answer : The volume of oxygen at STP is 112.0665 L
Solution : Given,
The number of moles of 
= 5 moles
At STP, the temperature is 273 K and pressure is 1 atm.
Using ideal gas law equation :
where,
P = pressure of gas
V = volume of gas
n = the number of moles
T = temperature of gas
R = gas constant = 0.0821 L atm/mole K (Given)
By rearranging the above ideal gas law equation, we get
Now put all the given values in this expression, we get the value of volume.
Therefore, the volume of oxygen at STP is 112.0665 L
First find the oxidation states of the various atoms:
<span>in Cr2O2 2- Cr @ +1; In NH3 N @ +3; in CrO3 Cr @ +3, N2 N @ 0 </span>
<span>Note that N gained electrons, ie, was reduced; Cr was oxidized </span>
<span>Now there is a problem, because B has NH4+ which the problem did not, and is not balanced, showing e- in/out </span>
<span>B.NH4+ → N2 </span>
<span>Which of the following is an oxidation half-reaction? </span>
<span>A.Sn 2+ →Sn 4+ + 2e- </span>
<span>Sn lost electrons so it got oxidized</span>
Answer : The equilibrium concentration of 
in the solution is, 
Explanation :
The dissociation of acid reaction is:
Initial conc. c 0 0
At eqm. c-x x x
Given:
c = 
The expression of dissociation constant of acid is:
![K_a=\frac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BC_6H_5COO%5E-%5D%7D%7B%5BC_6H_5COOH%5D%7D)
Now put all the given values in this expression, we get:
![6.3\times 10^{-5}=\frac{(x)\times (x)}{[(7.0\times 10^{-2})-x]}](https://tex.z-dn.net/?f=6.3%5Ctimes%2010%5E%7B-5%7D%3D%5Cfrac%7B%28x%29%5Ctimes%20%28x%29%7D%7B%5B%287.0%5Ctimes%2010%5E%7B-2%7D%29-x%5D%7D)
Thus, the equilibrium concentration of in the solution is,
I think it’s B 5.54 x 10^2g
Answer:
The amount of water converted from liquid to gas with 6,768 joules is approximately 3.035 g
Explanation:
The amount of heat required to convert a given amount of liquid to gas at its boiling point is known as the latent heat of evaporation of the liquid
The latent heat of evaporation of water, Δ
≈ 2,230 J/g
The relationship between the heat supplied, 'Q', and the amount of water in grams, 'm', evaporated is given as follows
Q = m × Δ
Therefore, the amount of water, 'm', converted from liquid to gas at the boiling point temperature (100°C), when Q = 6,768 Joules, is given as follows;
6,768 J = m × 2,230 J/g
∴ m = 6,768 J /(2,230 J/g) ≈ 3.035 g
The amount of water converted from liquid to gas with 6,768 joules = m ≈ 3.035 g.
