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3 years ago

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You need to produce a buffer solution that has a pH of 4.97. You already have a solution that contains 10. mmol (millimoles) of

acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution?

1 answer:

motikmotik3 years ago

5 0

Answer:

16.9 mmoles are needed to add to the solution

Explanation:

To solve this, we can apply the Henderson Hasselbach equation:

Acetic acid → CH₃COOH

CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka

That is the acid base equilibrium, to determine the amount:

pKa of acetic acid is 4.74

Henderson Hasselbach formula is:

pH = pKa + log (base/acid)

We replace → 4.97 = 4.74 + log (B / 10mmoles)

Let's verify mmoles of B

4.97 - 4.74 = log (B/10)

0.23 = log (B/10)

10^(0.23) = 10^(log (B/10)

1.698 = B/10

B = 1.698 . 10 → 16.9 mmoles

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An airplane travels 2100 km at 1000km/hE. It encounters a wind and slows to 800 km/h E for the next 1300 km. What is the average

Deffense [45]

Answer:

The average velocity of the airplane for this trip is 1684.21 km/h

Explanation:

Average velocity is the rate of change of displacement with time. That is,

Average velocity = $\frac{Displacement }{Change in time}$ = Δx / Δt = $\frac{x2 - x1}{t2 - t1}$

Now we will calculate the time taken by the airplane for the first motion before it encounters a wind.

From,

Velocity = $\frac{Distance traveled}{Time taken}$

Time = $\frac{Distance traveled}{Velocity}$

Therefore, Time = $\frac{2100km }{1000km/h}$

Time = 2.1h

This is the time taken before the airplane encounters a wind.

Hence, t1 = 2.1h

Now, For the time taken by the airplane when it encounters a wind

Also from,

Velocity = $\frac{Distance traveled}{Time taken}$

Time = $\frac{Distance traveled}{Velocity}$

Therefore, Time = $\frac{1300km }{800km/h}$

Time = 1.625h

Hence, t2 = 1.625h

Now, to calculate the average velocity

Average velocity = $\frac{x2 - x1}{t2 - t1}$

x1= 2100, x2= 1300, t1= 2.1h and t2= 1.625h

Hence, Average velocity = $\frac{1300 - 2100}{1.625 - 2.1}$

Average velocity = 1684.21 km/h

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3 years ago

Scientists use patterns in the periodic table to what?

Luba_88 [7]

I found this on google

“The periodic table is important because its is organized to provide a great deal of information about elements and how they relate to one another in one-easy-to-use reference. The table can be used to predict the properties of elements, even those that have not been discovered.”

I hope this helps

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3 years ago

Compare and contrast physical changes with chemical changes.

Alekssandra [29.7K]

There are several differences between a physical and chemical change in matter or substances. A physical change in a substance doesn't change what the substance is. In a chemical change where there is a chemical reaction, a new substance is formed and energy is either given off or absorbed.

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Many double-displacement reactions are enzyme-catalyzed via the "ping pong" mechanism, so called because the reactants appear to

zhenek [66]

Answer:

<u>D. It will decrease by a factor of 4</u>

Explanation:

According to the question , the equation follows :

$A+B\rightarrow C+D$

Rate law : This states the rate of reaction is directly proportional to concentration of reactants with each reactant raised to some power which may or may not be equal to the stoichiometeric coefficient.

$Rate\ \alpha [A]^{a}[B]^{b}$

$r=[A]^{a}[B]^{b}$ .................(1)

STEP": First, find out the power "a" and "b"

a+b = 3 (because it is given that the reaction follow 3rd order-kinetics)

According to question, <u><em>doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,</em></u>

r' = 2r if [A'] = 2[A]

Here [B] is uneffected means [B']=[B]

hence new rate =

$r'=[A']^{a}[B']^{b}$

Put the value of [A'] , [B'] and r' in the above equation:

$2r=[2A]^{a}[B]^{b}$ ...........(2)

Divide equation (1) by (2) we , get

$\frac{2r}{r}=\frac{[2A]^{2}[B]^{b}}{[A]^{a}[B]^{b}}$

$2= 2(\frac{A}{A})^{a}\times (\frac{B}{B})^{b}$

Here A and A cancel each other

B and B cancel each other

We get,

$2= 2^{a}\times 1^{b}$

1^b = 1 ( power of 1 = 1)

$2= 2^{a}$

This is possible only when a = 1

We know that : a + b = 3

1 + b = 3

b =3 -1 = 2

b = 2

Hence the rate law becomes :

$r=[A]^{a}[B]^{b}$

<u> $r=[A]^{1}[B]^{2}$ .............(3)</u>

Look in the question now, it is asked to calculate the concentration of [B],if cut in half

Hence

[B']=1/2[B]

Insert the value of [B'] in equation (3)

$r'=[A]^{1}[B']^{2}$

$r'=[A]^{1}(\frac{1}{2}[B])^{2}$

$r'=\frac{1}{4}[A]^{1}[B]^{2}$ ............(a)

But

$r=[A]^{a}[B]^{b}$ ..............(b)

Compare equation (a) and (b) , we get

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3 years ago

For the balanced equation shown below, if the reaction of 0.112 grams of

Alekssandra [29.7K]

Answer:

The answer is 74.5%.

Explanation:

As we know that % yield= $\frac{actual yield}{theoretical yield}$ x 100%.

Therefore,

Step 1 Calculate Theoretical yield:

0.112 $H_{2}$ x $\frac{1 mol H_{2} }{2.016 g of H{2} }$ x $\frac{4 mol H_{2}O }{4 mol H_{2} }$ x $\frac{18.02 g H_{2}O }{1 mol H_{2}O}$ = 1.001 g $H_{2}O$

Now Step 2

% yield = $\frac{actual yield}{theoretical yield}$ x 100% = $\frac{0.745g}{1.001g}$ = 74.5%

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