Answer:
% composition O = 19.9%
% composition Cu = 80.1%
Explanation:
Given data:
Total mass of compound = 3.12 g
Mass of copper = 2.50 g
Mass of oxygen = 3.12 - 2.50 = 0.62 g
% composition = ?
Solution:
Formula:
<em>% composition = ( mass of element/ total mass)×100</em>
% composition Cu = (2.50 g / 3.12 g)×100
% composition Cu = 0.80 ×100
% composition Cu = 80.1%
For oxygen:
<em>% composition = ( mass of element/ total mass)×100</em>
% composition O = (0.62 g / 3.12 g)×100
% composition O = 0.199 ×100
% composition O = 19.9%
Explanation:
To solve this problem, follow these steps;
- Obtain a balanced equation of the reaction and familiarize with the reactants and products.
- Find the number of moles of the reacting species since they are the known matter in terms of quantity.
- From the number of moles, determine the limiting reactant.
- The limiting reactant is the one given in short supply.
- Such reactant determines the extent of the reaction.
- Compare the moles of this specie to that of the products using the balanced equation.
- Obtain the mole of the desired product and find the mass or desired quantity.
- simply work from the known specie to the unknown
learn more:
Number of moles brainly.com/question/13064292
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Answer is: the percent by mass of NaHCO₃ is 2,43%.
m(NaHCO₃) = 10 g.
V(H₂O) = 400 ml.
d(H₂O) = 1 g/ml.
m(H₂O) = V(H₂O) · d(H₂O).
m(H₂O) = 400 ml · 1 g/ml.
m(H₂O) = 400 g.
m(solution) = m(H₂O) + m(NaHCO₃).
m(solution) = 400 g + 10 g.
m(solution) = 410 g.
ω(NaHCO₃) = 10 g ÷ 410 g · 100%.
ω(NaHCO₃) = 2,43 %
Answer:carbon-14 levels in the atmosphere
Explanation:
When carrying out radiocarbon dating, the level of carbon-14 in a sample is compared with the level of carbon 14 in the atmosphere because, objects exchange carbon-14 with the atmosphere.
Comparison of the activities of carbon-14 in the atmosphere and in the sample gives the age of the sample since the half-life of carbon-14 is a constant.
Answer:
A
Explanation:
Because it does not change entirely.
