Answer:

Explanation:
The expression for Clausius-Clapeyron Equation is shown below as:
Where,
P is the vapor pressure
ΔHvap is the Enthalpy of Vaporization
R is the gas constant (8.314×10⁻³ kJ /mol K)
c is the constant.
For two situations and phases, the equation becomes:

Given:
= 13.95 torr
= 144.78 torr
= 25°C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (25 + 273.15) K = 298.15 K
= 298.15 K
= 75°C = 348.15 K
So,





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>
Molarity is one of the method of expressing concentration of solution. Mathematically it is expressed as,
Molarity =

Given: Molarity of solution = 5.00 M
Volume of solution = 750 ml = 0.750 l
∴ 5 =

∴
number of moles = 3.75Answer: Number of moles of KOH present in solution is 3.75.
Answer:
Each Y atom needs three electrons to complete its octet by forming three bonds to hydrogen. There is one unshared pair of electrons and three bonding pairs of electrons. The bonds in the product are covalent.
Explanation:
Recall that group 5A elements contain five electrons in their outermost shell. These five electrons consists of a lone pair and three electrons that can form three bonds with hydrogen to give YH3 where Y is the group 5A element.
The YH3 molecule contains one lone(unshared) pair of electrons as well as three bonding pairs of electrons. The compounds are covalent.