The percentage adds up to 100%.
<h3>Percentages</h3>
Mass of mixture = mass of container+mixture - mass of container = 56.779 - 54.558 = 2.221 g
CuCO3 is insoluble in water. Thus:
Mass of CuCO3 = mass of beaker and residue - mass of beaker = 78.875 - 77.575 = 1.300 g
Mass of NaCl = mass of mixture - mass of CuCO3 = 2.221 - 1.300 = 0.921 g
%NaCl (w/w) = weight of NaCl/weight of mixture = 10.921/2.221 = 41.468%
% (w/w) CuCO3 = weight of CuCO3/weight of mixture = 1.300/2.221 = 58.532%
Addition of percentages = 41.468 + 58.532 = 100%
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Answer:
The easiest way is to titrate a sample with a solution of a base.
Explanation:
They include elements 57-71 (lanthanides) and 89-103 (actinides)
Answer:
The balanced equation is:
2 HNO3 + Mg ---> Mg(NO3)2 + H2
From the equation, we can see that we need twice the moles of HNO3 than the moles of Mg
Moles of Mg:
Molar mass of Mg = 24 g/mol
Moles = Given mass / Molar Mass
Moles of Mg = 4.47 / 24 = 0.18 moles (approx)
Hence, 2(moles of Mg) = 0.36 moles of HNO3 will be consumed
Number of moles of HNO3 after the reaction is finished is the number of unreacted moles of HNO3
Unreacted moles of HNO3 = Total Moles - Moles consumed
Unreacted moles of HNO3 = 0.64 moles (approx)
Since we approximated the value of moles of Mg, the value of remaining moles of HNO3 will also be approximate
From the given options, we can see that 0.632 moles is the closest value to our answer
Therefore, 0.632 moles will remain after the reaction
Answer:
A strong base produces more ions in solution than a weak base.
Explanation:
