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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)
Answer:
A) (3.2g)
Explanation:
Did you reposed this? Because I remember answering this
Answer:
Explanation:
mass % of C = 0.27/0.45*100 = 60%
mass % of H = 0.02/0.45*100 = 4.4%
mass % of O = 0.16/0.45*100 = 35.6%
Total = 60%+4.4%+ 35.6% = 100%
Answer:
c 43.38 g
Explanation:
The reaction between MnO2 and HCl can be represented by the following balanced equation:
MnO2 + HCl ---> Cl2 + MnCl2 + H2O
From the balanced equation, the theoretically required molar ratio of MnO2 to HCl is 1:1, therefore the yields would have been expected to be equal.
For the fact that HCl gives a higher yield (65.02g) than MnO2 (60.25g) according to the problem statement, HCl should be in excess, while the limiting reagent should be MnO2 .
Thus, the theoretical yield of Cl2 will be 60.25 g.
By definition, the percentage yield is given by
% Yield = (Actual Yield) / (Theoretical Yield),
This can be simplified to
Actual Yield = % Yield * Theoretical Yield
Plugging in the given values we have
Actual Yield = 72% * 60.25 = 43.38 g
