Number<span> of protons found in the nucleus of an </span>atom<span>.</span>
<u>Answer:</u> The ratio of the mass ratio of S to O in SO to the mass ratio of S to O in is 2.
<u>Explanation:</u>
We are given:
Mass of sulfur = 32 grams
Mass of oxygen = 16 grams
- <u>Taking SO molecule:</u>
A chemical compound having a chemical formula of SO
Mass of Sulfur in SO = 32 g
Mass of Oxygen in SO = 16 g
Mass ratio of Sulfur to oxygen in SO = .....(1)
- <u>Taking molecule:</u>
A chemical compound having a chemical formula of
Mass of Sulfur in = 32 g
Mass of Oxygen in = (2 × 16) = 32 g
Mass ratio of Sulfur to oxygen in = .....(2)
Taking ratio of 1 and 2, we get:
Hence, the ratio of the mass ratio of S to O in SO to the mass ratio of S to O in is 2.
Answer:
Mass = 234 g
Explanation:
Given data:
Mass of NaCl produced = ?
Moles of chlorine = 2.0 mol
Solution:
Chemical equation:
2Na + Cl₂ → 2NaCl
now we will compare the moles of NaCl with chlorine gas.
Cl₂ : NaCl
1 : 2
2.0 : 2/1×2.0 = 4.0 mol
Mass of NaCl;
Mass = number of moles × molar mass
Mass = 4.0 mol × 58.5 g/mol
Mass = 234 g
Explanation:
We have to calculate value.
It is known that at the equivalence point concentration of acid is equal to the concentration of anion formed.
Hence, [HA] =
Now, relation between and pH is as follows.
pH =
Putting the values into the above formula as follows.
pH =
4.23 = (as [HA] = )
= 4.23 (as log (1) = 0)
or, = 4
Thus, we can conclude that of given weak acid is 4.
Answer:- 27.2 g
Solution:- The balanced equation for the reaction of sulfuric acid with sodium bicarbonate is:
From this equation, one mole of sulfuric acid reacts with two moles of sodium bicarbonate. We could calculate the moles of sulfuric acid from it's given volume and molarity. Then using mol ratio, we could calculate the moles of sodium bicarbonate required. Moles could easily be converted to grams on multiplying by molar mass.
The calculations would be shown as:
=
So, 27.2 g of sodium bicarbonate are required to neutralize the given sulfuric acid solution.