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Darya [45]
3 years ago
13

(2pts) During the Purification of Lactate Dehydrogenase (LDH) experiment, you will need 50ml of buffer A150. Buffer A150 is 30mM

Tris (pH 8.8) and 150mM NaCl. Given 1L of 1.5M Tris and 500ml of 5M NaCl, how much of each stock do you need to make 250ml of the A150 buffer
Chemistry
1 answer:
torisob [31]3 years ago
6 0

Answer:

The answer is "20 \ mL"

Explanation:

Given:

Molarity= number of moles

because it is 1 Liter

\to \frac{0.03\ moles}{1.5 moles}=0.02\ L= 20 \ mL \ of\  Tris\\\\

therefore,

it takes 20 mL of Tris.

\to \frac{0.150 \ moles}{5\ moles} =0.03\ L\\\\

                     = 30 \ mL \ of\ Nacl

So, take 20 \ mL\ of\ NaCl.

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what is the concentration of a dextrose solution prepared by diluting 14 ml of a 1.0 m dextrose solution to 25 ml using a 25 ml
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The concentration of a dextrose solution prepared by diluting 14 ml of a 1.0 M dextrose solution to 25 ml using a 25 ml volumetric flask is 0.56M.

Concentration is defined as the number of moles of a solute present in the specific volume of a solution.

According to the dilution law, the degree of ionization increases on a dilution and it is inversely proportional to the square root of concentration. The degree of dissociation of an acid is directly proportional to the square root of a volume.

M₁V₁=M₂V₂

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10 months ago
If 72.5 grams of calcium metal (Ca) react with 65.0 grams of oxygen gas (O2) in a synthesis reaction, how many grams of the exce
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2Ca + O2 = 2CaO
First, determine which is the excess reactant
72.5 g Ca (1 mol) =1.8089725036
(40.078 g)

65 g O2 (1 mol) =2.0313769611
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Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.

Now do the stoichiometry
72.5 g Ca (1 mol Ca) (1 mol O2)
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