86.66 % by weight of ascorbic acid in the tablet, when 0. 1220 g vitamin c tablet was dissolved in acid. This required 11. 50 ml of 0. 01740 m 
to reach the endpoint.
<h3>What is a balanced chemical equation?</h3>
A balanced chemical reaction is an equation that has equal numbers of each type of atom on both sides of the arrow.
→ 
So, the balanced chemical equation for the reaction will be:
→ 
That means, 3 mol of of ascorbic acid reacts with 1 mol of
Moles of available is 0.0174 X 0.0115 =0.0002001 mol
Moles of ascorbic acid to be yielded should be 3 X 0.0002001 =0.0006003
So, percent mass of ascorbic acid in the tablet will be:
(0.1057 / 0.1220) X 100 %
=86.66 %
Hence, the third option is the correct answer.
Learn more about balanced chemical equations here:
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Answer:
Rubidium-85=61.2
Rubidium-87=24.36
Atomic Mass=85.56 amu
Explanation:
To find the atomic mass, we must multiply the masses of the isotope by the percent abundance, then add.
<u>Rubidium-85 </u>
This isotope has an abundance of 72%.
Convert 72% to a decimal. Divide by 100 or move the decimal two places to the left.
- 72/100= 0.72 or 72.0 --> 7.2 ---> 0.72
Multiply the mass of the isotope, which is 85, by the abundance as a decimal.
- mass * decimal abundance= 85* 0.72= 61.2
Rubidium-85=61.2
<u>Rubidium-87</u>
This isotope has an abundance of 28%.
Convert 28% to a decimal. Divide by 100 or move the decimal two places to the left.
- 28/100= 0.28 or 28.0 --> 2.8 ---> 0.28
Multiply the mass of the isotope, which is 87, by the abundance as a decimal.
- mass * decimal abundance= 87* 0.28= 24.36
Rubidium-87=24.36
<u>Atomic Mass of Rubidium:</u>
Add the two numbers together.
- Rb-85 (61.2) and Rb-87 (24.36)
Your answer is D. 8
16 = 2^4
72 = 2^3*3^2
So you'll choose 2^3 = 8
Concentration is the number of moles of solute in a fixed volume of solution
Concentration(c) = number of moles of solute(n) / volume of solution (v)
25.0 mL of water is added to 125 mL of a 0.150 M LiOH solution and solution becomes more diluted.
original solution molarity - 0.150 M
number of moles of LiOH in 1 L - 0.150 mol
number of LiOH moles in 0.125 L - 0.150 mol/ L x 0.125 L = 0.01875 mol
when 25.0 mL is added the number of moles of LiOH will remain constant but volume of the solution increases
new volume - 125 mL + 25 mL = 150 mL
therefore new molarity is
c = 0.01875 mol / 0.150 L = 0.125 M
answer is 0.125 M
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