CHEMISTRY COLLEGE

50 mL of 0.1 M acetic acid is mixed with 50 mL of 0.1 M sodium acetate (the conjugate base). The Ka of acetic acid is approximately 1. 74 X 10 -5. What is the pH of the resulting solution?

Answers

Answer 1

Answer:

Answer:

4.76

Explanation:

In this case, we have to start with the buffer system:

$CH_3COOH~->~CH_3COO^-~+~H^+$

We have an acid ( $CH_3COOH$ ) and a base ( $CH_3COO^-$ ). Therefore we can write the henderson-hasselbach reaction:

$pH~=~pKa+Log([CH_3COO^-])/([CH_3COOH])$

If we want to calculate the pH, we have to calculate the pKa:

$pH=-Log~Ka=4.76$

According to the problem, we have the same concentration for the acid and the base 0.1M. Therefore:

$[CH_3COO^-]=[CH_3COOH]$

If we divide:

$([CH_3COO^-])/([CH_3COOH])~=~1$

If we do the Log of 1:

$Log~1=~zero$

So:

$pH~=~pKa$

With this in mind, the pH is 4.76.

I hope it helps!

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Which of the following shows the combustion of a hydrocarbon?A. 2C2H2 +502 + 4CO2 + 2H20O B. CO2 + H2O → H2CO3O C. NaOH + HCl → NaCl + H20O D. C2H4 + Cl2 → C2H4Cl2
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To what extend must a soln conc 40mg 4g NO3 pa cm3 be dilute to yield one of conc 16mg 4g NO3 pa cm3
A student who is performing this experiment pours an 8.50 mL sample of the saturated borax solution into a 10 mL graduated cylinder after the borax solution had cooled to a certain temperature T. The student rinses the sample into a small beaker using distilled water, and then titrates the solution with a 0.500 M HCl solution. 12.00 mL of the HCl solution is needed to reach the endpoint of the titration.Calculate the value of Ksp for borax at temperature T.

Lewisite (2-chloroethenyldichloroarsine) was once manufactured as a chemical weapon, acting as a lung irritant and a blistering agent. During World War II, British biochemists developed an antidote which came to be known as British anti-Lewisite (BAL) (2,3-disulfanylpropan-1-ol). Today, BAL is used medically to treat toxic metal poisoning. Complete the reaction between Lewisite and BAL by giving the structure of the organic product and indicating the coefficient for the number of moles of HCl produced in the reaction.

Answers

Answer:

2 HCl

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

Consider the following reaction: Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq)If you react an excess of Pb(NO3)2with 26.3 g of NaCl, and you isolate 52.1 g of PbCl2, what is your percent yield?

Answers

Answer:

$\large \boxed{84.7 \, \%}$

Explanation:

Mᵣ: 58.44 278.11

Pb(NO₃)₂ + 2NaCl ⟶ PbCl₂ + 2NaNO₃

m/g: 26.3

1. Moles of NaCl

$\text{Moles of NaCl} = \text{26.3 g NaCl} * \frac{\text{1 mol NaCl}}{\text{58.44 g NaCl}} = \text{0.4505 mol NaCl}$

(b) Moles of PbCl₂

$\text{Moles of PbCl${_2}$} = \text{0.4505 mol NaCl} * \frac{\text{1 mol PbCl${_2}$}}{\text{2 mol NaCl}} = \text{0.2253 mol PbCl${_2}$}$

(c) Theoretical yield of PbCl₂

$\text{Mass of PbCl${_2}$} = \text{0.2253 mol PbCl${_2}$} * \frac{\text{278.11 g PbCl${_2}$}}{\text{1 mol PbCl${_2}$}} = \text{61.52 g PbCl${_2}$}$

(d) Percent yield

$\text{Percent yield} = \frac{\text{ actual yield}}{\text{ theoretical yield}} * 100 \,\% = \frac{\text{52.1 g}}{\text{61.52 g}} * 100 \, \% = \mathbf{84.7 \,\%}\n\n\text{The percent yield is $\large \boxed{\mathbf{84.7 \, \% }}$}$

Use the drop-down menus to answer the questions.Which organism begins this food chain and is a producer?
Which organism gains energy from eating the frog?
Which organism has the most available energy in this food chain?

Answers

Answer:

grass

snake

grass

Explanation:

just did it :)

Can you send a picture? That’s the only way I can see the problems

But, the producers should be on the bottom of the food chain, likely grass or something of the sort (plants)

If the frog has a line pointing to another organism, then that’s the organism that gains energy from eating it

Dumbledore thinks the students are now confident with significant figures; Snape is not so confident of this... Begrudgingly Snape decides to move forward in his lesson. His next lesson is in naming ionic compounds. He presents his students with sodium hydrogen sulfate and potassium permanganate. Which pair of ionic formulas is correct for these two compounds?(A) NaHSO3, KMnO4
(B) NaHSO2, KMnO4
(C) NaHSO4, KMnO4
(D) NaHSO4, KMnO3
(E) NaHSO3, KMnO3

Answers

Answer:

C) $NaHSO_(4), KMnO_(4)$

Explanation:

- For the sodium hydrogen sulfate:

The ending -ate of the word sulfate indicates that the compound comes from the ion sulfate that is $SO_(4)^(-)$ , so the compound formed by this ion will be $NaHSO_(4)$

- For the potassium permanganate:

The ending -ate of the word permanganate indicates that the compound comes from the ion $MnO^(-)_(4)$ , so the compound formed by this ion will be $KMnO_(4)$

How many grams of solid sodium cyanide should be added to 1.00 L of a 0.119 M hydrocyanic acid solution to prepare a buffer with a pH of 8.809

Answers

Answer:

1.62 g

Explanation:

Given that:

Concentration of HCN = 0.119 M

Assuming the ka 4.00 × 10⁻¹⁰

The pKa of HCN (hydrocyanic acid) = -log (Ka)

= - log ( 4.00 × 10⁻¹⁰)

= 9.398

pH of buffer = 8.809

Using Henderson Hasselbach equation:

$pH = pKa + log ([conjugate\ base ])/(acid)$

$pH = pKa + log ([CN^-])/([HCN])$

$8.809 = 9.398 +log ([CN^-])/([HCN])$

$log ([CN^-])/([HCN])= 8.809 - 9.398$

$log ([CN^-])/([HCN])= -0.589$

$([CN^-])/([HCN])= 0.2576$

[CN^-] = 0.2576[HCN]

[CN^-] = 0.2756 (0.119) L

[CN^-] = 0.033 M

∴

The amount of NaCN (sodium cyanide) is calculated as follows:

$= 1.00 L * (0.033 \ mol \ NacN )/(1 \ L ) * (49.01 \ g)/(1 \ mol \ of \ NacN)$

= 1.62 g

Which of the following is an example of physical change?

Answers

Answer:

The glass cup falling from the counter

Explanation:

the glass isn't changing in any chemical way. it's still made of the same material, just broken apart.

Final answer:

Physical changes involve the alteration of the state or appearance of matter, without changing the composition. An example is solid wax turning into liquid wax when heated, or steam condensing inside a cooking pot.

Explanation:

The question asks for an example of a physical change. Physical changes involve alterations in the state or appearance of matter, without changing its composition. For example, solid wax turning into liquid wax when heated is a physical change. The wax is still the same substance, it's just in a different state. Similarly, steam condensing inside a cooking pot is also a physical change. The water vapor turns back into liquid water, but it's still water. These are distinguished from chemical changes, which transform one substance into a different substance.

Learn more about Physical Change here:

brainly.com/question/17931044

#SPJ2

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To measure the solubility product of lead (II) iodide (PbI2) at 25°C, you constructed a galvanic cell that is similar to what you used in the lab. The cell contains a 0.5 M solution of a lead (II) nitrate in one compartment that connects by a salt bridge to a 1.0 M solution of potassium iodide saturated with PbI2 in the other compartment. Then you inserted two lead electrodes into each half-cell compartment and closed the circuit with wires. What is the expected voltage generated by this concentration cell? Ksp for PbI2 is 1.4 x 10-8. Show all calculations for a credit.