Answer:
"1.4 mL" is the appropriate solution.
Explanation:
According to the question,
Now,
Increase in volume will be:
⇒ 
By putting the given values, we get
Answer:
the spotlight effect.
Explanation:
The spotlight effect is a tendency to think that people get noticed more often than they really do. It is an overestimation of the situation regarding the concern of getting observed. It concerns the self-confidence of an individual. For example, an individual feels that everybody in a party would notice him for a bad pair of shoes while in reality, it does not concern them.
As per the question, the overestimation of people's reaction is known as the spotlight effect.
Answer:
Theoretical yield of the reaction = 34 g
Excess reactant is hydrogen
Limiting reactant is nitrogen
Explanation:
Given there is 100 g of nitrogen and 100 g of hydrogen
Number of moles of nitrogen = 100 ÷ 28 = 3·57
Number of moles of hydrogen = 100 ÷ 2 = 50
Reaction between nitrogen and hydrogen yields ammonia according to the following chemical equation
N2 + 3H2 → 2NH3
From the above chemical equation for every mole of nitrogen that reacts, 3 moles of hydrogen will be required and 2 moles of ammonia will be formed
Now we have 3·57 moles of nitrogen and therefore we require 3 × 3·57 moles of hydrogen
⇒ We require 10·71 moles of hydrogen
But we have 50 moles of hydrogen
∴ Limiting reactant is nitrogen and excess reactant is hydrogen
From the balanced chemical equation the yield will be 2 × 3·57 moles of ammonia
Molecular weight of ammonia = 17 g
∴ Theoretical yield of the reaction = 2 × 3·57 × 17 = 121·38 g
I would say CuSO4 or Copper Sulfate, as option 1 is Methane and would create a fire so heat, and the last one is Sugar which doesn't conduct electricity. And C6H6 I believe is not soluble in water.<span />
Answer : The correct option is, (C) 1.1
Solution : Given,
Initial moles of 
= 1.0 mole
Initial volume of solution = 1.0 L
First we have to calculate the concentration .
The given equilibrium reaction is,
Initially c 0
At equilibrium 
The expression of 
will be,
![K_c=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
where,
= degree of dissociation = 40 % = 0.4
Now put all the given values in the above expression, we get:
Therefore, the value of equilibrium constant for this reaction is, 1.1
