Hihi!
The correct answer is B. The mass number of this nuclide is 131! It's proton number is 56 and it has 75 neutrons (131 - 56 = 75 neutrons).
I hope I helped!
-Jailbaitasmr
Answer:
The answer to the question is
The temperature at which the vapor pressure will be 5.00 times higher than it was at 331 K is 353.0797 K.
Explanation:
To solve the question, we make use of the Clausius-Clapeyron equation as follows

Where P₁ = Initial pressure
P₂ = Final pressure
T₁ = Initial temperature = 331 K
T₂ = Final temperature
dvapH = ΔvapH = Heat of vaporization = 70.83 kJ / mol.
R = Universal gas constant = 8.3145. J K⁻¹ mol⁻¹
We are required to find the temperature when P₂ = 5 × P₁
Therefore we have
=
or T₂ =
= 353.0797 K
The vapor pressure be 5.00 times higher than it was at 331 K when the temperature is raised to 353.0797 K.
Answer:
The partial pressure of nitrogen is 597.5 torr
Explanation:
Step 1: Data given
Mol fraction of N = 0.781
Mol fraction of O = 0.209
Mol fraction Ar = 0.010
Atmospheric pressure = 765.0 torr
The partial pressure is given by Pi = Xi * Pt
⇒ χ
i = the mole fraction of gas i in the mixture
⇒ P
total = the total pressure of the mixture
Step 2: Calculate the partial pressure of nitrogen
P(N) = mol fraction of N * atmospheric pressure
P(N) = 0.781 * 765.0 torr = 597.5 torr
The partial pressure of nitrogen is 597.5 torr
As the atomic number increases the atomic radius increases since it's adding more electrons to the outermost shell