An experiment requires 0.025 l of a 0.5 m glucose solution. you have 84 ml of a 20 % (m/v) glucose stock solution. how would you prepare the required concentration?

Concentration of the given stock solution of glucose = 20% (m/v)

It means that 20 g glucose is present in 100 mL solution

Volume of the 20 % solution given is 84 mL

Moles of glucose =

Moles of glucose in 84 mL solution:

Calculating molarity of 84 mL of 20 % glucose solution:

Finding the volume of this solution required to prepare 0.0250L of 0.5 M glucose solution:

hurricanes form in warm water

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2