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Semmy [17]
4 years ago
13

The pka of the α-carboxyl group of glutamine is 2.17, and the pka of its α-amino group is 9.13. calculate the average net charge

on glutamine if it is in a solution that has a ph of 8.20. if the charge is positive, do not enter a \" \" sign.
Chemistry
2 answers:
IgorLugansk [536]4 years ago
4 0
I think u add the numbers and get the total and divide by the set of numbers to get the average.
Serjik [45]4 years ago
3 0

Answer:

-0.1051

Explanation:

First, we need to find the degree of dissociation (α) of each of the groups using the Handerson-Halsebach equation:

pH = pKa + log (α/1-α)

log (α/1-α) =  pH - pKa

log (1-α/α) = pKa - pH

1-α/α =10^{pKa - pH}

1-α = 10^{pKa - pH}*α

α = \frac{1}{10^{pKa-pH} +1}

So, fo the α-carboxyl group:

α = \frac{1}{10^{-6.03} +1}

α = 1.0000

And for the α-amino group

α = \frac{1}{10^{0.93} +1}

α = 0.1051

In the ionization, the α-carboxyl group goes from neutral to a negative ion with charge -1, and the α-amino group goes from a positive ion with charge +1 to a neutral compound. So, the net charge is how many ions are presented multiplied by the charge.

For the first one, the value of α is the value of ions, but for the second, α represents the neutral, so the ions are 1-α:

-1*(1.000) + (+1)*(1-0.1051) = -0.1051

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Nataliya [291]
Quantitative means to count how many of something. The correct answer is B, It has two eyes. Hope this helps
6 0
3 years ago
Read 2 more answers
Can someone solve this for me I'm confused.
Artemon [7]

Answer:

310.53 g of Cu.

Explanation:

The balanced equation for the reaction is given below:

CuSO₄ + Zn —> ZnSO₄ + Cu

Next, we shall determine the mass of CuSO₄ that reacted and the mass Cu produced from the balanced equation. This can be obtained as follow:

Molar mass of CuSO₄ = 63.5 + 32 + (16×4)

= 63.5 + 32 + 64

= 159.5 g/mol

Mass of CuSO₄ from the balanced equation = 1 × 159.5 = 159.5 g

Molar mass of Cu = 63.5 g/mol

Mass of Cu from the balanced equation = 1 × 63.5 = 63.5 g

Summary:

From the balanced equation above,

159.5 g of CuSO₄ reacted to produce 63.5 g of Cu.

Finally, we shall determine the mass of Cu produced by the reaction of 780 g of CuSO₄. This can be obtained as follow:

From the balanced equation above,

159.5 g of CuSO₄ reacted to produce 63.5 g of Cu.

Therefore, 780 g of CuSO₄ will react to produce = (780 × 63.5)/159.5 = 310.53 g of Cu.

Thus, 310.53 g of Cu were obtained from the reaction.

6 0
3 years ago
Solid carbon is known to react with oxygen gas to produce carbon dioxide. If a mass of 1.2g of carbon is burned in oxygen, 4.4g
Elden [556K]

Now ,

C + O2  → CO2

According to above equation, 1 mole of carbon reacts with one mole of oxygen to produce one mole of carbon dioxide.Thus this implies that 12 g of carbon reacts with 32 g of O2 to produce 44 g of CO2.

No of moles = mass of the substance/molecular mass of the substance.

In this case 1.2 g of carbon reacts with "x "g of O2 to produce 4.4 g of CO2.

No of moles of carbon in this case = 1.2÷ 12 = 0.1 moles.

No of moles of carbon dioxide formed = 4.4÷44 =0.1 moles

Thus already discussed above, 1 mole of carbon reacts with 1 mole of oxygen to produce 1 mole of carbon dioxide. Hence to produce 0.1 mole of CO2 ,0.1 mole of carbon needs to react with 0.1 mole of oxygen.

Also number of moles of O2 = mass of O2÷ molar mass of O2

Substituting number of moles of O2 as 0.1 we get

mass of O2(x)  = Number of moles of O2 × Molar mass of O2

Mass of O2 (x) = 0.1 × 32= 3.2 g

Thus mass of 3.2 g O2 reacts with 1.2 g of CO2 to produce 4.4 g of CO2.



7 0
3 years ago
A car reaches 80 km/hr, then suddenly crashed into a concrete wall. Without a seat
Debora [2.8K]

Answer:

C

Explanation:

inertia

inertia is the reluctance of an object at rest or moving to change its state of rest or motion

because the car was moving, the dummy moves with the speed of the car..so when the crash happened the dummy continued to move..it was feeling reluctant to come to rest so was moving

4 0
3 years ago
A 50.0-ml sample of 0.50 m hcl is titrated with 0.50 m naoh. what is the ph of the solution after 28.0 ml of naoh have been adde
hram777 [196]

The pH of the solution after addition of 28 mL of NaOH is added to HCl is \boxed{{\text{0}}{\text{.85}}} .

Further Explanation:

The proportion of substance in the mixture is called concentration. The most commonly used concentration terms are as follows:

1. Molarity (M)

2. Molality (m)

3. Mole fraction (X)

4. Parts per million (ppm)

5. Mass percent ((w/w) %)

6. Volume percent ((v/v) %)

Molarity is a concentration term that is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by M and its unit is mol/L.

The formula to calculate the molarity of the solution is as follows:

{\text{Molarity of solution}}=\dfrac{{{\text{Moles}}\;{\text{of}}\;{\text{solute}}}}{{{\text{Volume }}\left({\text{L}} \riht){\text{ of solution}}}}          

                             ......(1)        

                         

Rearrange equation (1) to calculate the moles of solute.

{\text{Moles}}\;{\text{of}}\;{\text{solute}}=\left( {{\text{Molarity of solution}}}\right)\left({{\text{Volume of solution}}}\right)       ......(2)

Substitute 0.50 M for the molarity of solution and 50 mL for the volume of solution in equation (2) to calculate the moles of HCl.

\begin{aligned}{\text{Moles}}\;{\text{of}}\;{\text{HCl}}&= \left({{\text{0}}{\text{.50 M}}}\right)\left( {{\text{50 mL}}} \right)\left( {\frac{{{\text{1}}{{\text{0}}^{ - 3}}{\text{ L}}}}{{{\text{1 mL}}}}} \right)\\&= 0.02{\text{5 mol}}\\\end{aligned}

Substitute 0.50 M for the molarity of solution and 28 mL for the volume of solution in equation (2) to calculate the moles of NaOH.

\begin{aligned}{\text{Moles}}\;{\text{of}}\;{\text{NaOH}}&=\left( {{\text{0}}{\text{.50 M}}} \right)\left( {{\text{28 mL}}} \right)\left( {\frac{{{\text{1}}{{\text{0}}^{ - 3}}{\text{ L}}}}{{{\text{1 mL}}}}}\right)\\&= 0.014{\text{ mol}}\\\end{aligned}

The reaction between HCl and NaOH occurs as follows:

{\text{NaOH}} + {\text{HCl}} \to {\text{NaCl}} + {{\text{H}}_2}{\text{O}}

The balanced chemical reaction indicates that one mole of NaOH reacts with one mole of HCl. So the amount of remaining HCl can be calculated as follows:

\begin{aligned}{\text{Amount of HCl remaining}}&= 0.02{\text{5 mol}} - 0.01{\text{4 mol}}\\&= {\text{0}}{\text{.011 mol}} \\\end{aligned}

The volume after the addition of NaOH can be calculated as follows:

\begin{aligned}{\text{Volume of solution}} &= {\text{50 mL}} + {\text{28 mL}}\\&= {\text{78 mL}}\\\end{aligned}

Substitute 0.011 mol for the amount of solute and 78 mL for the volume of solution in equation (1) to calculate the molarity of new HCl solution.

\begin{aligned}{\text{Molarity of new HCl solution}}&= \left({{\text{0}}{\text{.011 mol}}} \right)\left( {\frac{1}{{{\text{78 mL}}}}}\right)\left( {\frac{{{\text{1 mL}}}}{{{{10}^{ - 3}}\;{\text{L}}}}} \right)\\&= 0.1410{\text{2 M}}\\&\approx {\text{0}}{\text{.141 M}}\\\end{aligned}

pH:

The acidic strength of an acid can be determined by pH value. The negative logarithm of hydronium ion concentration is defined as pH of the solution. Lower the pH value of an acid, the stronger will be the acid. Acidic solutions are likely to have pH less than 7. Basic or alkaline solutions have pH more than 7. Neutral solutions have pH equal to 7.

The formula to calculate pH of an acid is as follows:

{\text{pH}}=- {\text{log}}\left[ {{{\text{H}}^ + }}\right]     ......(3)

Here,

\left[{{{\text{H}}^ + }}\right] is hydrogen ion concentration.

HCl is a strong acid so it dissociates completely. So the concentration of   also becomes 0.141 M.

Substitute 0.141 M for \left[{{{\text{H}}^ + }}\right] in equation (3).

\begin{aligned}{\text{pH}}&= - {\text{log}}\left({0.141} \right)\\&=0.85\\\end{aligned}

So the pH of the solution is 0.85.

Learn more:

1. Which indicator is best for titration between HI and  ? brainly.com/question/9236274

2. Why is bromophenol blue used as an indicator for antacid titration? brainly.com/question/9187859

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Acid-base titrations

Keywords: molarity, pH, HCl, NaOH, 0.85, 0.141 M, moles of HCl, moles of NaOH, 50 mL, 0.50 M, 28 mL, 0.025 mol, 0.014 mol, 0.011 mol, 78 mL.

4 0
3 years ago
Read 2 more answers
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