C. atoms of reactants rearrange to form new substances
The m/z and relative abundance of the ions contributed to the peak at 21.876 min. The relative abundance will be 21.876%.
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What is relative abundance?</h3>
- The proportion of atoms with a particular atomic mass present in an element sample taken from a naturally occurring sample is known as the relative abundance of an isotope.
- When the relative abundances of an element's isotopes are multiplied by their atomic masses and the results are added up, the result is the element's average atomic mass, which is a weighted average.
- Chemists often divide the number of atoms in a particular isotope by the sum of the atoms in all the isotopes of that element, then multiply the result by 100 to determine the percent abundance of each isotope in a sample of that element.
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Answer:
A- Energy from sunlight becomes glucose during photosynthesis
Explanation:
The law of conservational photosynthetic energy states that atoms cannot be created nor destroyed; the atoms can only be rearranged into new molecules and ions. This implies that the enzymes in photosynthesis are aided by energy to transform into glucose, which the plant needs.
The solution is 45.7 % (NH₄)₂CO₃ by mass.
Mass of solution = 0.0332 kg + 0.0395 kg = 0.0727 kg
% (NH₄)₂CO₃ = Mass of (NH₄)₂CO₃/Total mass × 100 % = 0.0332 kg/0.0727 kg × 100 % = 45.7 %
Answer:
pH = 12.15
Explanation:
To determine the pH of the HCl and KOH mixture, we need to know that the reaction is a neutralization type.
HCl + KOH → H₂O + KCl
We need to determine the moles of each compound
M = mmol / V (mL) → 30 mL . 0.10 M = 3 mmoles of HCl
M = mmol / V (mL) → 40 mL . 0.10 M = 4 mmoles of KOH
The base is in excess, so the HCl will completely react and we would produce the same mmoles of KCl
HCl + KOH → H₂O + KCl
3 m 4 m -
1 m 3 m
As the KCl is a neutral salt, it does not have any effect on the pH, so the pH will be affected, by the strong base.
1 mmol of KOH has 1 mmol of OH⁻, so the [OH⁻] will be 1 mmol / Tot volume
[OH⁻] 1 mmol / 70 mL = 0.014285 M
- log [OH⁻] = 1.85 → pH = 14 - pOH → 14 - 1.85 = 12.15