Answer:
The volume of a sample of gas (2.49 g) was 752 mL at 1.98 atm and 62∘C 62 ∘ C .
The first step in the scientific methods is ask a question
Oxygen and glucose and energy. <span />
Both sides of the equation would be equal.
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<span>LiOH+HBr---> LiBr +h20. Moles of LiOH = 10/24 = 0.41moles. According to stoichiometry, moles of LiOH = moles of LiBr = 0.41moles. Therefore mass of LiBr =moles of LiBr x molecular weight of LiBr = o.41 x 87 = 35.67g. Hope it helps </span>
