Explanation:
The given data is as follows.
50 ml of
, 50 ml of 
And, it is known that at STP 1 mole of a gas occupies 22.4 L. Hence, moles present in 50 ml of gas are as follows.
(As 1 L = 1000 ml)
=
moles
So, according to the given equation
moles of
reacts with
moles of
.
Hence, moles of
is equal to the moles of
and
.
Therefore, moles of
=
moles
1 mole of
= 22.4 L
moles =
= 50 ml of product
Thus, we can conclude that 50 ml of products if pressure and temperature are kept constant.
Answer: 6.71 g
Explanation: 



Limiting reagent is the reagent which limits the formation of product. Excess reagent is one which is in excess and thus remains unreacted.
Thus lithium is the limiting reagent and nitrogen is the excess reagent.
As can be seen from the balanced chemical equation, 6 moles of lithium reacts with 1 mole of nitrogen to give 2 moles of lithium nitride.
Thus 0.578 moles of lithium react with 0.096 moles of nitrogen.
6 moles of lithium give = 2 moles of lithium nitride
Thus 0.578 moles of lithium give=
of lithium nitride.
Mass of lithium nitride 
Mass of lithium nitride
=
Answer:
Explanation:
There are 7 protons in nitrogen
What is the oxidation number of iodine in kio4? -7?
According to Henderson–Hasselbalch Equation,
pH = pKa + log [Acetate] / [Acetic Acid]
As,
pKa = -log Ka
pKa = -log (1.8 × 10⁻⁵)
pKa = 4.74
So,
pH = 4.74 + log [Acetate] / [Acetic Acid]
4.34 = 4.74 + log [Acetate] / [Acetic Acid]
4.34 - 4.74 = log [Acetate] / [Acetic Acid]
-0.40 = log [Acetate] / [Acetic Acid]
Taking Antilog on both sides,
[Acetate] / [Acetic Acid] = 0.398