The mass of sodium hydrogen carbonate : 10.5 g
<h3>Further explanation</h3>
Given
1.5 dm' of CO₂
1 mol gas= 24 L at RTP(25 °C, 1 atm)
Required
the mass of sodium hydrogen carbonate
Solution
Decomposition reaction of Sodium hydrogen carbonate :
2 NaHCO₃ (s) ⇒ Na₂
CO₃ (s) + H₂
O(g) + CO₂ (g)
mol CO₂ :
From the equation, mol ratio of NaHCO₃ : CO₂ (g) = 2 : 1, so mol NaHCO₃ :
Mass NaHCO₃(MW=23+1+12+3.16=84 g/mol) :
Lower temperature
Let's verify
- Pressure=P
- volume=V
- Temperature=T
As per Boyles law
As per Charles law
So
At higher altitudes lower the pressure so lower the temperature
We use the gas law named Charle's law for the calculation of the second temperature. The law states that,
V₁T₂ = V₂T₁
Substituting the known values,
(0.456 L)(65 + 273.15) = (3.4 L)(T₁)
T₁ = 45.33 K
Boyle’s Law illustrates the inverse relationship of volume and pressure. It follows the formula p1V1 = P2V2 , where P1V1 denotes initial pressure and volume and P2V2 denotes values of pressure and volume.
Now, let us work out for what is asked above.
a. if the pressure is doubled
50.0 p = V x 2p
V = 50.0 p / 2p
= 50.0 /2
= 25.0 m^3
b. if the pressure is cut in half
50.0 p = V x p/2
100 p = V x p
V = 100 m^3
c. if the pressure is tripled
50.0 p = V x 3p
V = 50.0 p / 3p
= 50.0 /3
=16.7 m^3
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Answer:
The molarity of the dissolved NaCl is 6.93 M
Explanation:
Step 1: Data given
Mass of NaCl = 100.0 grams
Volume of water = 100.0 mL = 0.1 L
Remaining mass NaCl = 59.5 grams
Molar mass NaCl= 58.44 g/mol
Step 2: Calculate the dissolved mass of NaCl
100 - 59. 5 = 40.5 grams
Step 3: Calculate moles
Moles NaCl = 40.5 grams / 58.44 g/mol
Moles NaCl = 0.693 moles
Step 4: Calculate molarity
Molarity = moles / volume
Molarity dissolved NaCl = 0.693 moles / 0.1 L
Molarity dissolved NaCl = 6.93 M
The molarity of the dissolved NaCl is 6.93 M
