Answer:
The partial pressure of the other gases is 0.009 atm
Explanation:
Step 1: Data given
Air is about 78.0% nitrogen molecules and 21.0% oxygen molecules and 1% of other gases.
The atmospheric pressure = 0.90 atm
Step 2: Calculate mol fraction
If wehave 100 moles of air, 78 moles will be nitrogen,
21 moles will be oxygen, and 1 mol will be other gases.
Mol fraction = 1/100 = 0.01
Step 3: Calculate the partial pressure of the other gases
Pgas = Xgas * Ptotal
⇒ Pgas = the partial pressure = ?
⇒ Xgas = the mol fraction of the gas = 0.01
⇒Ptotal = the total pressure of the pressure = 0.90 atm
Pgas = 0.01 * 0.90 atm
Pgas = 0.009 atm
The partial pressure of the other gases is 0.009 atm
Answer:
Draw circles to represent the electron shell of each atom overlapping the circles where the atoms are bonded. Add dots to represent the outer electrons of one type of atom (H). Add crosses to represent the outer electrons of the other type of atom (Cl). Make sure the electrons are always in pairs.
Answer:
2HgS + 3O2 → 2HgO + 2SO2
The coefficients are: 2, 3, 2, 2
Explanation:
HgS + O2 → HgO + SO2
The equation can be balance as follow:
Put 3 in front of O2 as shown below:
HgS + 3O2 → HgO + SO2
Now we can see that there are 6 atoms of O on the left side of the equation and a total of 3 atoms on the right side. It can be balance by putting 2 in front of HgO and SO2 as shown below:
HgS + 3O2 → 2HgO + 2SO2
Now we have 2 atoms of both Hg and S on the right side and 1atom each on the left. It can be balance by putting 2 in front of HgS as shown below:
2HgS + 3O2 → 2HgO + 2SO2
Now the equation is balanced.
The coefficients are: 2, 3, 2, 2
The law of conservation of mass(matter) states that matter(mass) can neither be created nor destroyed during a chemical reaction but changes from one form to another. An unbalanced equation suggests that matter has been created or destroyed. While a balanced equation proofs that matter can never be created but changes to different form. This is the more reason we have count the atoms of an element on both side of the equation to see if they are balanced irrespective of the new form they assume in the product
<h2>Natural Abundance for 10B is 19.60%</h2>
Explanation:
- The natural isotopic abundance of 10B is 19.60%.
- The natural isotopic abundance of 11B is 80.40%.
- The isotopic masses of boron are 10.0129 u and 11.009 u respectively.
For calculation of abundance of both the isotopes -
Supposing it was 50/50, the average mass would be 10.5, so to increase the mass we need a more percentage of 11.
Determining it as an equation -
10x + 11y= 10.8
x+y=1 (ratio)
10x + 10y = 10
By taking the denominator away from the numerator
we get;
y = 0.8
x + y = 1
∴ x = 0.2
To get percentages we need to multiply it by 100
So, the calculated abundance is 80% for 11 B and 20% 10 B.
