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Describe, in detail, how to make 5.5 liters of a 2.5 molar HCl solution from a 10.5 molar HCl stock solution.

Misha Larkins [42]

The stock solution contains 10.5 moles of HCl per litre. A 5.5 litre solution of 2.5M HCl contains 5.5x2.5 = 13.75moles of HCl. Since every litre of stock solution provides 10.5M HCl, the amount of stock solution needed is 13.75/10.5 = 1.309L. Therefore you would dilute 1.309L of stock solution to 5.5L

5 0

4 years ago

Why is it permissible to use a wet bottle when first obtaining your kmno4 solution?

Murrr4er [49]

For the answer to the question above, well presumably because the exact concentration of the composition KMnO4 solution doesn't matter. If the concentration of the KMnO4 solution is important (usually in titrations etc.) then it is not allowed to use a wet bottle. The water in the bottle will dilute the KMnO4 solution and change the concentration of the said compound.

5 0

3 years ago

The reaction for photosynthesis producing glucose sugar and oxygen gas is:

Anvisha [2.4K]

1.5 grams of glucose is produced from 2.20 g of CO₂.

To find the mass of glucose produced, first you must know the balanced reaction. For this, the Law of Conservation of Matter is followed.

The law of conservation of matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.

So, in this case, the balanced reaction is:

6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the amounts of moles of each reactant and product participate in the reaction:

CO₂: 6 moles
H₂O: 6 moles
C₆H₁₂O₆: 1 mole
O₂: 6 moles

So, you know that 2.20 g of CO₂ react, whose molar weight is 44.01 g/mole. By definition of molar mass, 1 mole of CO₂ has 44.01 g. So, the number of moles that 2.20 grams of the compound represent is calculated as:

$moles of CO_{2} =2.20 grams*\frac{1 mole}{44.01 grams}$

moles of CO₂= 0.05 moles

Now you must follow the following rule of three: if by stoichiometry of the reaction 6 moles of CO₂ produce 1 mole of C₆H₁₂O₆, 0.05 moles of CO₂ produce how many moles of C₆H₁₂O₆?

$moles of C_{6} H_{12} O_{6} =\frac{0.05moles of CO_{2} *1 mole of C_{6} H_{12} O_{6}}{6moles of CO_{2}}$