How many milliliters of a 3.0 M HCL solution are required to make 250.0 milliliters of 1.2 M HCL

A 5.00-mL sample of blood was treated with trichloroacetic acid to precipitate proteins. After centrifugation, the resulting sol

Anarel [89]

Explanation:

According to the Beer-lambert' s law,

Absrobance = absrobptitvity × lenght × concentration

So, if we will plot a graph between absrobance and concentration then we will obtain a straight line

Hence, formula to calculate slope of the graph is as follows.

The slope of graph = $\frac{\text{difference in absrobance at two points}}{\text{difference in concentrations at those points}}$

The given data is as follows.

$A_{1}$ = 0.412 (first point)

$A_{2}$ = 0.642 (second point)

$C_{1}$ = 0.240 ppm (first point)

$C_{2}$ = 0.475 ppm (second point)

Hence, putting these values into the above formula we will calculate the value of slope as follows.

Slope = $\frac{\text{difference in absrobance at two points}}{\text{difference in concentrations at those points}}$

= $\frac{0.642 - 0.412}{0.475 - 0.240}$

= $\frac{0.23}{0.235}$

= 0.978

It is known that for the graph, line equation can be written as follows.

y = mx + c ........... (1)

where, C = intercept

m = slope

Hence, calculate the value of intercept as follows.

0.412 = $0.24 \times 0.978$ + c

c = 0.178

As it is given that the absorbance values 0.454 (the y-axis value). Therefore, putting this value into equation (1) we get the following.

y = mx + c

0.454 = $x \times 0.978$ + 0.178

0.276 = 0.978x

x = 0.282 ppm

Thus, we can conclude that the concentration of Pb in the given sample is 0.282 ppm.

3 0

3 years ago

Suppose that 25.0 mL of 0.10 M CH3COOH (aq) is titrated with 0.10 M NaOH (aq). What is the pH after the addition of 10.0 mL of 0

olya-2409 [2.1K]

Answer:

$pH=-1.37$

Explanation:

We are given that 25 mL of 0.10 M $CH_3COOH$ is titrated with 0.10 M NaOH(aq).

We have to find the pH of solution

Volume of $CH_3COOH=25mL=0.025 L$

Volume of NaoH=0.01 L

Volume of solution =25 +10=35 mL= $\frac{35}{1000}=0.035 L$

Because 1 L=1000 mL

Molarity of NaOH=Concentration OH-=0.10M

Concentration of H+= Molarity of $CH_3COOH$ =0.10 M

Number of moles of H+=Molarity multiply by volume of given acid

Number of moles of H+= $0.10\times 0.025$ =0.0025 moles

Number of moles of $OH^-=0.10\times 0.01$ =0.001mole

Number of moles of H+ remaining after adding 10 mL base = 0.0025-0.001=0.0015 moles

Concentration of H+= $\frac{0.0015}{0.035}=4.28\times 10^{-2} m/L$

pH=-log [H+]=-log [4.28 $\times 10^{-2}$ ]=-log4.28+2 log 10=-0.631+2

$pH=-1.37$

6 0

2 years ago

In acid solution, water can add to the double bond of 2-butenedioic acid to form 2-hydroxysuccinic acid.

satela [25.4K]

So, what is the question?

6 0

2 years ago

Please someone help ill give out the brainliest I will pleaseee just help

snow_lady [41]

Answer:

have a great time at y avg be have a great time at all today I was going to answer restroom for spring break is over and over again and again and again and again and again and again your hhhha

5 0

2 years ago

Base changes phenolphthalein to pink it is true or false

galben [10]

Answer:

It’s false

Explanation:

it could be true if the question mentioned alkaline solution

4 0

2 years ago

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