3 H2SO4 + 2 Al(OH)3 → Al2(SO4)3 + 6 H2O
(2.14 g Al(OH)3) / (78.0036 g Al(OH)3/mol) x (3 mol H2SO4 / 2 mol Al(OH)3) / (0.210 mol/L H2SO4) =
0.19596 L = 196 mL H2SO4
Answer:
36
Explanation:
Since the sample was undiluted the number of colonies is the number that grew on the nutrient agar which is 36 colonies. If it was diluted for example let say 0.1 ml from a dilution in which 1 ml of the sample was added to 9 ml of water, and it grew colonies then 0.1 ml yielded 6 colonies, 1 ml of the diluted sample will yield 60 colonies and 10 ml will have 600 colonies and therefore the 1 ml undiluted sample will have 600 colonies.
Answer: Servsafe 9
Explanation:
Remove food from the surface
clean the surface
rinse the surface
sanitize the surface
allow the surface to air dry
Answer:
110ml
Explanation:
<em>Using the dilution equation, C1V1 = C2V2</em>
<em>Where C1 is the initial concentration of solution</em>
<em>C2 is final concentration of solution</em>
<em>V1 is intital volume of solution</em>
<em>V2 is final volume of solution.</em>
From the question , C1=6M, C2=0.5M, V1=10ml, V2=?



volume of water added = final volume -initial volume
= 120-10
=110ml