The m/z and relative abundance of the ions contributed to the peak at 21.876 min. The relative abundance will be 21.876%.
<h3>
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:
Plants add water to the atmosphere
Explanation:
Answer : The value of 
at this temperature is 66.7
Explanation : Given,
Pressure of 
at equilibrium = 0.348 atm
Pressure of 
at equilibrium = 0.441 atm
Pressure of 
at equilibrium = 10.24 atm
The balanced equilibrium reaction is,
The expression of equilibrium constant for the reaction will be:
Now put all the values in this expression, we get :
Therefore, the value of at this temperature is 66.7
For example, copper is used for electrical<span> wiring because it is a </span>good conductor of electricity<span>. </span>Metal<span> particles are held together by strong metallic bonds, which is why they have high melting and boiling points. The free electrons in </span>metals<span> can move through the </span>metal<span>, allowing </span>metals<span> to conduct </span>electricity<span>.</span>
Answer:
1.99grams
Explanation:
- First, we need to calculate the molar mass of the compound: Ca(HCO3)2
Ca = 40g/mol, H = 1g/mol, C = 12g/mol, O = 16g/mol
Hence, Ca(HCO3)2
= 40 + {1 + 12 + 16(3)}2
= 40 + {13 + 48}2
= 40 + {61}2
= 40 + 122
= 162g/mol
Molar mass of Ca(HCO3)2 = 162g/mol
- Next, we calculate the mass of oxygen in one mole of the compound, Ca(HCO3)2.
Oxygen = {16(3)}2
= 48 × 2
= 96g of Oxygen
- Next, we calculate the percentage composition of oxygen by mass by dividing the mass of oxygen in the compound by the molar mass of the compound i.e.
% composition of O = 96/162 × 100
= 0.5926 × 100
= 59.26%.
- The number of moles of the compound, Ca(HCO3)2, must be converted to mass by using the formula; mole = mass/molar mass
0.0207 = mass/162
Mass = 162 × 0.0207
Mass = 3.353grams
However, in every gram of Ca(HCO3)2, there is 0.5926 g of oxygen
Hence, in 3.353grams of Ca(HCO3)2, there will be;
0.5926 × 3.353
= 1.986
= 1.99grams.
Therefore, there is 1.99grams of Oxygen in 0.0207 moles (3.353g) of Ca(HCO3)2.
