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tensa zangetsu [6.8K]
3 years ago
7

g Suppose you are titrating an acid of unknown concentration with a standardized base. At the beginning of the titration, you re

ad the base titrant volume as 1.94 mL. After running the titration and reaching the endpoint, you read the base titrant volume as 23.82 mL. What volume of base was required for the titration
Chemistry
1 answer:
melamori03 [73]3 years ago
8 0

Answer:

21.88mL is the volume of base required for the titration.

Explanation:

For an acid-base titration trying to find the concentration of an acid, you must add a known quantity of the acid and titrate it with an standarized base.

If you know the moles of base you add to the acid solution, these moles are equal to moles of acid.

In the buret of the titration, initial volume is 1.94mL and final volume is 23.82mL. The volume you are adding is the difference between initial and final volume, that is:

23.82mL - 1.94mL

<h3>21.88mL is the volume of base required for the titration.</h3>
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Answer:

Kc of reaction is 20.

Explanation:

The two proteins are X and Y.

The [X] = 1mM

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At equilibrium, [X] = 0.2mM  [Y] = 0.2mM

we know that equilibrium constant is:

Kc=\frac{[Products]}{[Reactants]}=\frac{[XY]}{[X][Y]}

[XY]= 1-0.20=0.80 mM

putting values:

Kc=\frac{0.80}{0.20X0.20}=20

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4 years ago
How much heat is required to warm 1.50L of water from 25.0C to 100.0C? (Assume a density of 1.0g/mL for the water.)
Masteriza [31]

<u>Answer:</u> The amount of heat required to warm given amount of water is 470.9 kJ

<u>Explanation:</u>

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Density of water = 1 g/mL

Volume of water = 1.50 L = 1500 mL    (Conversion factor:  1 L = 1000 mL)

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{1500mL}\\\\\text{Mass of water}=(1g/mL\times 1500mL)=1500g

To calculate the heat absorbed by the water, we use the equation:

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