how many milliliters of a stock solution of 2.00M KNO3 would you need to prepare 100.0mL of 0.150M KNO3?

The pH of an acid solution is 5.82. Calculate the Ka for the monoprotic acid. The initial acid concentration is 0.010 M.

blagie [28]

Answer:

The answer is

<h3> $Ka = 2.29 \times {10}^{ - 14} moldm^{ - 3}$ </h3>

Explanation:

The Ka of an acid when given the pH and concentration can be found by

where

c is the concentration of the acid

From the question

pH = 5.82

c = 0.010 M

Substitute the values into the above formula and solve for Ka

We have

Multiply through by - 2

Find antilog of both sides

We have the final answer as

<h3> $Ka = 2.29 \times {10}^{ - 14} moldm^{ - 3}$ </h3>

Hope this helps you

8 0

3 years ago

which of the following amino acids are used to measure the protein concentration by UV? A. His B. Pro C. Trp D. Phe E. Tyr

Goryan [66]

Answer:

C. Trp D. Phe E. Tyr

Explanation:

The concentration of a protein has a direct relation with absorbance of the protein in a UV spectrophotometer. The formula which relates concentration with absorbance is described as under:

A = ∈ x c x l

where, A = Absorbance

∈ = Molar extinction co-efficient

c = Concentration of absorbing species i.e. protein

l = Path length of light

Tryptophan (Trp), phenylalanine (Phe ) and tyrosine (Tyr) are three aromatic amino acids which are used to measure protein concentration by UV. It is mainly because of tryptophan (Trp), protein absorbs at 280 nm which gives us an idea of protein concentration during UV spectroscopy.

The table depicting the wavelength at which these amino acids absorb and their respective molar extinction coefficient is as under:

Amino acid Wavelength Molar extinction co-efficient (∈)

Tryptophan 282 nm 5690

Tyrosine 274 nm 1280

Phenylalanine 257 nm 570

In view of table above, we can easily see that Molar extinction co-efficient (∈) of Tryptophan is highest amongst all these 3 amino acids that is why it dominates while measuring concentration.

7 0

3 years ago

Choose two ways carbon returns to nature from animals

SVEN [57.7K]

Answer:

its D and C I did the test

8 0

3 years ago

Be sure to answer all parts. The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calcu

pochemuha

Answer : The volume of $CO_2$ will be, 514.11 ml

Explanation :

The balanced chemical reaction will be,

$HCO_3^-+HCl\rightarrow Cl^-+H_2O+CO_2$

First we have to calculate the mass of $HCO_3^-$ in tablet.

$\text{Mass of }HCO_3^-\text{ in tablet}=32.5\% \times 3.79g=\frac{32.5}{100}\times 3.79g=1.23175g$

Now we have to calculate the moles of $HCO_3^-$ .

Molar mass of $HCO_3^-$ = 1 + 12 + 3(16) = 61 g/mole

$\text{Moles of }HCO_3^-=\frac{\text{Mass of }HCO_3^-}{\text{Molar mass of }HCO_3^-}=\frac{1.23175g}{61g/mole}=0.0202moles$

Now we have to calculate the moles of $CO_2$ .

From the balanced chemical reaction, we conclude that

As, 1 mole of $HCO_3^-$ react to give 1 mole of $CO_2$

So, 0.0202 mole of $HCO_3^-$ react to give 0.0202 mole of $CO_2$

The moles of $CO_2$ = 0.0202 mole

Now we have to calculate the volume of $CO_2$ by using ideal gas equation.

$PV=nRT$

where,

P = pressure of gas = 1.00 atm

V = volume of gas = ?

T = temperature of gas = $37^oC=273+37=310K$

n = number of moles of gas = 0.0202 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get :

$(1.00atm)\times V=0.0202 mole\times (0.0821L.atm/mole.K)\times (310K)$

$V=0.51411L=514.11ml$

Therefore, the volume of $CO_2$ will be, 514.11 ml

6 0

2 years ago

Aspirin can be made in the laboratory by reacting acetic anhydride (c4h6o3) with salicylic acid (c7h6o3) to form aspirin (c9h8o4

Temka [501]

V(C₄H₆O₃) = 5.00 mL.
d(C₄H₆O₃) = 1.08 g/mL.
m(C₄H₆O₃) = V(C₄H₆O₃) · d(C₄H₆O₃).
m(C₄H₆O₃) = 5.00 mL · 1.08 g/mL.
m(C₄H₆O₃) = 5.4 g.
n(C₄H₆O₃) = m(C₄H₆O₃) ÷ M(C₄H₆O₃).
n(C₄H₆O₃) = 5.4 g ÷ 102 g/mol.
n(C₄H₆O₃) = 0.0529 mol.
n(C₇H₆O₃) = 2.08 g ÷ 138.1 g/mol.
n(C₇H₆O₃) = 0.015 mol; limiting reactant.
From chemical reaction: n(C₄H₆O₃) : n(C₉H₈O₄) = 1 : 1.
n(C₉H₈O₄) = 0.015 mol.
m(C₉H₈O₄) = 0.015 mol · 180.16 g/mol.
m(C₉H₈O₄) = 2.71 g; theoretical yield.
percent yield od aspirine = 2.57 g ÷ 2.71 g · 100% = 94.83%.

7 0

3 years ago