Answer : The vapor pressure of propane at is 17.73 atm.
Explanation :
The Clausius- Clapeyron equation is :
where,
= vapor pressure of propane at = ?
= vapor pressure of propane at normal boiling point = 1 atm
= temperature of propane =
= normal boiling point of propane =
= heat of vaporization = 24.54 kJ/mole = 24540 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:
Hence, the vapor pressure of propane at is 17.73 atm.
We are going to use Avogadro's constant to calculate how many molecules of
carbons dioxide exist in lungs:
when 1 mole of CO2 has 6.02 x 10^23 molecules, so how many molecules in
CO2 when the number of moles is 5 x 10^-2
number of molecules = moles of CO2 * Avogadro's number
= 5 x 10^-2 * 6.02 x 10^23
= 3 x 10^22 molecules
∴ There are 3 x 10^22 molecules in CO2 exist in lungs
<h3>
Answer:</h3>
2 L Ne
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Using Dimensional Analysis
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
0.07 mol Ne (g)
<u>Step 2: Identify Conversions</u>
STP - 22.4 L per mole
<u>Step 3: Convert</u>
- Set up:
- Multiply:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 1 sig fig.</em>
1.568 L Ne ≈ 2 L Ne
The moles of ammonium nitrate needed to dissolve 0.35 moles
The moles of water that will react is 0.35 moles as due to ratio
so mass of water will be 0.35 x 18=6.3g
MASS OF WATER WILL BE 6.3 g