The average atomic mass of your mixture is 1.03 u
.
The average atomic mass of H is the weighted average of the atomic masses of its isotopes.
We multiply the atomic mass of each isotope by a number representing its relative importance (i.e., its % abundance).
Thus,
0.99 × 1.01 u = 0.998 u
0.002 × 2.01 u = 0.004 u
0.008 × 3.02 u = <u>0.024 u</u>
TOTAL = 1.03 u
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<span>-Boron (B)
-Silicon (Si)
-Germanium (Ge)
-Arsenic (As)
-Antimony (Sb)
-Tellurium (Te)
-Polonium (Po)
<span>-Astatine (At)</span></span>
Answer:
C + 2H2 ⇒ CH4
Explanation:
In order to balance a chemical equation you need to make sure that the number of atoms on both sides are equal
C + H2 = CH4
C = 1
H = 2
Products:
C = 1
H = 4
H2 = 2 × 2 = 4
C + 2H2 ⇒ CH4
Hope this helps.
Considering the definition of percentage by mass, the mass percentage of CaCO₃ is 68.59%.
<h3>What is mass percentage</h3>
The percentage by mass expresses the concentration and indicates the amount of mass of solute present in 100 grams of solution.
In other words, the percentage by mass of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.
The percentage by mass is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:
<h3>Mass percentage of CaCO₃</h3>
In this case, you know:
- mass of CaCO₃: 2.62 grams
- mass of limestone: 3.82 grams
Replacing in the definition of mass percentage:
<u><em>mass percentage= 68.59 %</em></u>
Finally, the mass percentage of CaCO₃ is 68.59%.
Learn more about percentage by mass:
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There are two ways to solve this problem. We can use the ICE method which is tedious and lengthy or use the Henderson–Hasselbalch equation. This equation relates pH and the concentration of the ions in the solution. It is expressed as
pH = pKa + log [A]/[HA]
where pKa = - log [Ka]
[A] is the concentration of the conjugate base
[HA] is the concentration of the acid
Given:
Ka = 1.8x10^-5
NaOH added = 0.015 mol
HC2H3O2 = 0.1 mol
NaC2H3O2 = 0.1 mol
Solution:
pKa = - log ( 1.8x10^-5) = 4.74
[A] = 0.015 mol + 0.100 mol = .115 moles
[HA] = .1 - 0.015 = 0.085 moles
pH = 4.74 + log (.115/0.085)
pH = 4.87