Answer:
Atomic number = 10
Mass number = 20
Explanation:
Mass number = neutrons+protons
Mass number = 10+10 = 20
The balloon will reach its maximum volume and it will burst.
Given:
- A weather balloon at sea level, with gas at 65.0 L volume, 745 Torr pressure, and 25C temperature.
- When the balloon was taken to an altitude at which temperature was 25C and pressure was 0.066atm its volume expanded.
- The maximum volume of the weather balloon is 835 L.
To find:
Whether the weather balloon will reach its maximum volume or not.
Solution:
The pressure of the gas in the weather balloon at sea level = 
The volume of the weather balloon at sea level = 
The temperature of the gas in the weather balloon at sea level:
The balloon rises to an altitude.
The pressure of the gas in the weather balloon at the given altitude:
The volume of the weather balloon at the given altitude = 
The temperature of the gas in the weather balloon at the given altitude:
Using the Combined gas law:
The maximum volume of the weather balloon= V = 835 L
The volume of the weather balloon at a given altitude is greater than its maximum volume which means the balloon will reach its maximum volume and it will burst.
Learn more about the combined gas law:
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1) Calcium carbonate contains 40.0% calcium by weight.
M(CaCO₃)=100.1 g/mol
M(Ca)=40.1 g/mol
w(Ca)=40.1/100.1=0.400 (40.0%)!
2) Mass fraction of this is excessive data.
3) The solution is:
m(Ca)=1.2 g
m(CaCO₃)=M(CaCO₃)*m(Ca)/M(Ca)
m(CaCO₃)=100.1g/mol*1.2g/40.1g/mol=3.0 g
Answer:
pHe = 3.2 × 10⁻³ atm
pNe = 2.5 × 10⁻³ atm
P = 5.7 × 10⁻³ atm
Explanation:
Given data
Volume = 1.00 L
Temperature = 25°C + 273 = 298 K
mHe = 0.52 mg = 0.52 × 10⁻³ g
mNe = 2.05 mg = 2.05 × 10⁻³ g
The molar mass of He is 4.00 g/mol. The moles of He are:
0.52 × 10⁻³ g × (1 mol / 4.00 g) = 1.3 × 10⁻⁴ mol
We can find the partial pressure of He using the ideal gas equation.
P × V = n × R × T
P × 1.00 L = 1.3 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K
P = 3.2 × 10⁻³ atm
The molar mass of Ne is 20.18 g/mol. The moles of Ne are:
2.05 × 10⁻³ g × (1 mol / 20.18 g) = 1.02 × 10⁻⁴ mol
We can find the partial pressure of Ne using the ideal gas equation.
P × V = n × R × T
P × 1.00 L = 1.02 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K
P = 2.5 × 10⁻³ atm
The total pressure is the sum of the partial pressures.
P = 3.2 × 10⁻³ atm + 2.5 × 10⁻³ atm = 5.7 × 10⁻³ atm
Answer:
The final pressure of the gas mixture after the addition of the Ar gas is P₂= 2.25 atm
Explanation:
Using the ideal gas law
PV=nRT
if the Volume V = constant (rigid container) and assuming that the Ar added is at the same temperature as the gases that were in the container before the addition, the only way to increase P is by the number of moles n . Therefore
Inicial state ) P₁V=n₁RT
Final state ) P₂V=n₂RT
dividing both equations
P₂/P₁ = n₂/n₁ → P₂= P₁ * n₂/n₁
now we have to determine P₁ and n₂ /n₁.
For P₁ , we use the Dalton`s law , where p ar1 is the partial pressure of the argon initially and x ar1 is the initial molar fraction of argon (=0.5 since is equimolar mixture of 2 components)
p ar₁ = P₁ * x ar₁ → P₁ = p ar₁ / x ar₁ = 0.75 atm / 0.5 = 1.5 atm
n₁ = n ar₁ + n N₁ = n ar₁ + n ar₁ = 2 n ar₁
n₂ = n ar₂ + n N₂ = 2 n ar₁ + n ar₁ = 3 n ar₁
n₂ /n₁ = 3/2
therefore
P₂= P₁ * n₂/n₁ = 1.5 atm * 3/2 = 2.25 atm
P₂= 2.25 atm
