In order to calculate the number of atoms present in 0.53 grams of P₂O₅, first calculate the number of moles for given mass.
As, Moles is given as,
Moles = Mass / M.mass
Moles = 0.53 g ÷ 283.88 g.mol⁻¹
Moles = 0.00186 moles
Now calculate the number of Molecules present in calculated moles of P₂O₅.
As,
1 Mole of P₂O₅ contains = 6.022 × 10²³ Molecules
So,
0.00186 Moles of P₂O₅ contain = X Molecules
Solving for X,
X = (0.00186 mol × 6.022 × 10²³ Molecules) ÷ 1 mol
X = 1.12 × 10²¹ Molecules
Also, in P₂O₅ there are 7 atoms, in 1.12 × 10²¹ Molecules there will be.....
= 1.12 × 10²¹ × 7
= 7.84 × 10²¹ Atoms
Result:
7.84 × 10²¹ Atoms are present in 0.53 grams of P₂O₅.
C.
centi- is essentially 10^2 of one meter.
If you had 100m, multiplying 100 by 10^2 (or 100) would give you 10000 cm.
Answer: Baking soda
Explanation:
Sodium bicarbonate has a chemical formula of NaHCO3, it is a white precipitate formed when excess carbon dioxide is bubbled through a concentrated solution of sodium hydroxide.
It is used in cooking and serves as baking powder.
Answer:
A loss of gain of electron.
Answer:
The answer to your question is V2 = 18.45 l
Explanation:
Data
Volume 1 = V1 = 12.3 l
Temperature 1 = T1 = 40°K
Volume 2 = V2 = ?
Temperature 2 = T2 = 60°K
Process
To solve this problem use Charles' law
V1/T1 = V2/T2
-Solve for V2
V2 = V1T2 / T1
-Substitution
V2 = (12.3 x 60) / 40
-Simplification
V2 = 738 / 40
-Result
V2 = 18.45 l
