(3) loses one electron and becomes positively charged
Lithium has one valence electron and Bromine has seven. Therefore Lithium will give up its one to Bromine for both to have an octet
1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N. Details about number of atoms can be found below.
How to calculate number of atoms?
The number of atoms of a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.
However, the number of moles of oxygen in glycine can be calculated using the following expression:
Molar mass of C₂H5O2N = 75.07g/mol
Mass of oxygen in glycine = 32g/mol
Hence; 32/75.07 × 7.51 = 3.2grams of oxygen in glycine
Moles of oxygen = 3.2g ÷ 16g/mol = 0.2moles
Number of atoms of oxygen = 0.2 × 6.02 × 10²³ = 1.205 × 10²³ atoms
Therefore, 1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N.
Learn more about number of atoms at: brainly.com/question/8834373
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The last set which is n=4 l=3 m=3 is a valid set
D. The number increases and then decreases for noble gases
