I would start by converting g/mL to g/L
75.00g/1000.0mL = 75.00g/L
Stoichiometrically, flip 75 onto the bottom so that grams cancel out and we are left with the number of L required.
(L/75.00g)(15g) -> this is essentially dividing 15g by 75g, which cancels the unit g, leaving us with 0.2L. If the question requires an answer in mL, just multiple the number of L by 1000.
The patient requires 200mL of glucose solution to receive his 15g of glucose.
It is highly reactive and when it is kept in open it does react with the oxygen present in the surroundings and burns therefore it is kept immersed in kerosene and please thank me and if you need more comment
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