Moles= mass divided by molar mass
Molar mass= 12.01(4) + 1.01(10)
= 58.14g/mol
Moles=14.5g / 58.14g/mol
=0.249
Therefore there are approx 0.249 moles in a 14.5g sample of C4H10
Answer:
Explanation:
<u>1. Energy to heat the liquid water from 55ºC to 100ºC</u>
<u>2. Energy to change the liquid to steam at 100ºC</u>
- n = 10.1g / 18.015g/mol = 0.5606mol
<u>3. Total energy</u>
As per the given chemical formula- Na2CO3.10H2O, one mole of the chemical compound contains 13 moles of oxygen atoms. Hence
Number of moles of oxygen atoms in one mole of Na2CO3.10H2O = 13
number of moles of oxygen atoms in 0.2 moles of Na2CO3.10H2O = 13 X 0.2 = 2.6
Now, one mole of a substance contains 6.022 X 10^23 particles of the substance. Thus
number of atoms of oxygen in one mole of oxygen atom = 6.022 X 10^23
number of moles of oxygen atoms in 2.6 moles of oxygen atoms = 2.6 X 6.022 X 10^23 = 15.657 X 10^23
= 1.566 X 10^24
Thus, there are 1.566 X 10^24 atoms of oxygen in 0.2 moles of Na2CO3.10H2O.
4.7
It would be 9*10 to the -3
Given:
M = 0.0150 mol/L HF solution
T = 26°C = 299.15 K
π = 0.449 atm
Required:
percent ionization
Solution:
First, we get the van't Hoff factor using this equation:
π = i MRT
0.449 atm = i (0.0150 mol/L) (0.08206 L atm / mol K) (299.15 K)
i = 1.219367
Next, calculate the concentration of the ions and the acid.
We let x = [H+] = [F-]
[HF] = 0.0150 - x
Adding all the concentration and equating to iM
x +x + 0.0150 - x = <span>1.219367 (0.0150)
x = 3.2905 x 10^-3
percent dissociation = (x/M) (100) = (3.2905 x 10-3/0.0150) (100) = 21.94%
Also,
percent dissociation = (i -1) (100) = (</span><span>1.219367 * 1) (100) = 21.94%</span>
