Question

ANSWER 5TH QUESTION
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Answer 1

$Question:\\\\How\:\: many\:\: atoms\:\: are\:\: present\:\: in \:\:28.4g\:\:of\:\:Cl \:\:atoms$

$Answer:\\Molar\:\:mass\:\:of\:\:Cl = 35.5$

$No.\:\:of\:\:moles\:\:of\:\:Cl\:\:in\:\:28.4g\:\:= \:\:\frac{28.4}{35.5}= \:\:0.8\:\:moles$

$one\:\:mole\:\:of\:\:Cl \:\:atoms\:\:have\:\:6.022\times10^2^3 \:\:atoms\\ \\ Then,\:\:0.8\:\:moles\:\:will\:\:have:\:\:0.8\times6.022\times10^2^3=\:\:4.8176\times10^2^3\:\:atoms\\\\Please \:\:mark\:\:me\:\:brainliest$