To calculate atomic mass, you have to take to weighted average of the isotopes' masses. What that means is M = RA*106 + (1 – RA)*104, where RA is relative abundance expressed in decimal form. If you simplify the right side of that equation, you get M = 2*RA + 104. Doing a little more algebra yields RA = (M –104)/2 = (104.4 – 104)/2 = 0.4 / 2 = 0.2, which is 20%. So the answer is B.
You multiply the number by 1000 when you convert a measurement from kilometres to metres.
The prefix <em>kilo</em> means <em>×1000</em>, so
1 <em>kilo</em>metre means 1 metre <em>×1000</em>.
Answer:
The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Explanation:
Given;
CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol
From the combustion reaction above, it can be observed that;
1 mole of methane (CH₄) released 890 kilojoules of energy.
Now, we convert 59.7 grams of methane to moles
CH₄ = 12 + (1x4) = 16 g/mol
59.7 g of CH₄ 
1 mole of methane (CH₄) released 890 kilojoules of energy
3.73125 moles of methane (CH₄) will release ?
= 3.73125 moles x -890 kJ/mol
= -3320.81 kJ
Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
173.99cm multiply 68.5 x 2.54 always use that number when converting inches to cm
<h3><u>Answer</u>;</h3>
≈ 4.95 g/L
<h3><u>Explanation;</u></h3>
The molar mass of KCl = 74.5 g/mole
Therefore; 0.140 moles will be equivalent to ;
= 0.140 moles × 74.5 g/mole
= 10.43 g
Concentration in g/L
= mass in g/volume in L
= 10.43/2.1
= 4.9667
<h3> <u> ≈ 4.95 g/L</u></h3>
