Formula of hydrated sodium carbonate : Na₂CO₃.10H₂O, so moles of water in one mole of hydrated salt = 10
<h3>Further explanation</h3>
Hydrate is a compound that binds water (H₂O), usually in the form of crystals/ solids
If these compounds are dissolved in water or heated, the hydrates can decompose:
Example: X.YH₂O (s) → X (aq) + YH₂O (l)
The formula for the hydrated compound contains: YH2O
The mole ratio shows the ratio of the coefficients of the hydrate compound
10.45 hydrated sodium carbonate(Na₂CO₃.xH₂O) were heated until 3.87 of 3.87of anhydrous (Na₂CO₃) remained, so
mass H₂O released :
mass Na₂CO₃ = 3.87 g
mol ratio Na₂CO₃(MW= g/mol) : H₂O(MW=18 g/mol) =
Answer:
34.2 g is the mass of carbon dioxide gas one have in the container.
Explanation:
Moles of 
:-
Mass = 49.8 g
Molar mass of oxygen gas = 32 g/mol
The formula for the calculation of moles is shown below:
Thus,
Since pressure and volume are constant, we can use the Avogadro's law as:-
Given ,
V₂ is twice the volume of V₁
V₂ = 2V₁
n₁ = ?
n₂ = 1.55625 mol
Using above equation as:
n₁ = 0.778125 moles
Moles of carbon dioxide = 0.778125 moles
Molar mass of 
= 44.0 g/mol
Mass of = Moles × Molar mass = 0.778125 × 44.0 g = 34.2 g
<u>34.2 g is the mass of carbon dioxide gas one have in the container.</u>
The answer:
we should know the meaning of each abbreviation:
ms means millisecond, its value is 10^-3 s
ns means means nanosecond, its value is 10^-9 s
ps means picosecond, its value is 10^-12 s
fs means femtosecond, its value is 1x 10^15 s
<span>Expressions of the quantity 556.2 x 10^-12 are</span>
556.2 x 10^-12 =556.2 ps
556.2 x 10^-12 =556.2 x 10^-9 x 10^-3= 556.2 x 10^-9 ms
556.2 x 10^-12 = 556.2 x 10^-3 x 10^-9 = 556.2 x 10^-3 ns
556.2 x 10^-12 = 556.2 x 10^- 27 x 10^15 = 556.2 x 10^- 27 fs
The volume of a balloon f a gas at 842 mm Hg and -23 celsius if it’s volume is 915 milliliters at a pressure of 1170 mm Hg And a temperature of 24 celsius is 0.22 litres
Explanation:
Data given:
Initial volume of the balloon having gas V1= 915ml OR 0.195 L
initial pressure of the gas P1= 1170 mm Hg OR 1.53 atm
initial temperature of the gas T1 = 24 celsius or 273.15 + 24 = 297.15 K
Final pressure of the gas P2 = 842 mm Hg or 1.10 atm
final temperature of the gas T2 = -23 degrees or 273.15 - 23 = 250.15 K
Final volume at final temperature and pressure V2=?
The formula used is of Gas Law:
= 
V2 = 
putting the values in the equation:
V2 = 
V2 = 0.22 litres is the volume
The volume is 0.22 litres at a pressure of 1170 mmHg and temperature of -23 degrees.
The mass number goes on top, and the atomic number goes on bottom.
Therefore, the answer is C:
40
K
19
