We know,

For given reaction, 

For , 2.41 moles of
:

We know :

Hence, this is the required solution.
Answer: % error of observation is 4.77%
Explanation:
Given:
Observation value = 415nm
theoretical value= 435.8nm
Percent error of observation = theoretical value- observation value/ theoretical value x 100 %
= 435.8-415/435.8= 0.04772 x 100 = 4.77%
therefore % error of observation is 4.77%
Molecule is a chemical substance that cannot be broken down into another chemical substance.
Answer:

Explanation:
Hello!
In this case, since the decomposition of sodium hydrogen carbonate is:

Thus, since there is a 2:1 mole ratio between the sodium hydrogen carbonate and sodium carbonate, and the molar masses are 84.01 and 105.99 g/mol respectively, we obtain the following theoretical yield:

Best regards!
Answer:
There are 6.022 × 1023 atoms of potassium in every mole of potassium. Since one mole of KOH contains one mole of K, the answer is 6.022×1023 atoms of K.
Explanation: