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makvit [3.9K]
4 years ago
9

You need to produce a buffer solution that has a pH of 5.26. 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
Chemistry
1 answer:
Andreas93 [3]4 years ago
3 0

The question is incomplete, complete question is :

You need to produce a buffer solution that has a pH of 5.50. 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? The pka of acetic acid is 4.74.

Answer:

33.11 millimoles of acetate we will need to add to this solution.

Explanation:

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[salt]}{[acid]})

Where :

tex]pK_a[/tex] = negative logarithm of acid dissociation constant of acid

[salt] = Concentration of salt

[Acid] = Concentration of salt

We have:

pH = 5.26

pK_a=4.74

[salt] =[CH_3COO^-] = ?

[acid] = [CH_3COOH]=10.0 mmol

5.26=4.74+\log(\frac{[CH_3COO^-]}{[10.0 mmol]})

[CH_3COO^-]=33.11 mmol

33.11 millimoles of acetate we will need to add to this solution.

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Answer:

They blow away from poles to the equator.

Explanation:

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In this case, we must take into account that global wind systems are formed by the constant increase in the temperature of the Earth’s surface. Thus, they drive the oceans’ surface currents. In such a way, we can say wind is the basic movement of air from an area of higher pressure to an area of lower pressure, for that reason they blow away from the poles to the equator.

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4 years ago
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Answer:

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3 years ago
The sum of the first 10 terms of an arithmetic progression is 120 and the sum of first twenty is 840. find sum of first 30 terms
STatiana [176]

Answer:

The sum of first 30 terms of the arithmetic progression is <u>2160.</u>

Explanation:

For an arithmetic progression, the sum of first n terms with first term as a and common difference d is given as:

S_n=\frac{n}{2}(2a+(n-1)d)

Now, it is given that:

For\ n=10,S_n=120\\For\ n=20,S_n=840

Now, plug in these values and frame two equations in a\ and\ d

S_{10}=\frac{10}{2}(2a+(10-1)d)\\120=5(2a+9d)\\2a+9d=\frac{120}{5}\\2a+9d=24------------1

S_{20}=\frac{20}{2}(2a+(20-1)d)\\840=10(2a+19d)\\2a+19d=\frac{840}{10}\\2a+19d=84-----------2

Now, we solve equations (1) and (2) for a\ and\ d. Subtract equation (1) from equation (2). This gives,

2a+19d-2a-9d=84-24\\19d-9d=60\\10d=60\\d=\frac{60}{10}=6

Now, plug in the value of d=6 in equation (1) and solve for a.

2a+9(6)=24\\2a+54=24\\2a=24-54\\2a=-30\\a=\frac{-30}{2}=-15

Plug in the values of a=-15,\ n=30\ and\ d=6 in the sum formula to find the sum of first 30 terms.

Now, the sum of first 30 terms is given as:

S_{30}=\frac{30}{2}(2(-15)+(30-1)(6))\\S_{30}=15(-30+29(6))\\S_{30}=15(-30+174)\\S_{30}=15(144)=2160

Therefore, the sum of first 30 terms of the arithmetic progression is 2160.

4 0
3 years ago
Chromium-51 is a radioisotope that is used to assess the lifetime of red blood cells The half-life of chromium-51 is 27.7 days.
8090 [49]

Answer:

11.9g remains after 48.2 days

Explanation:

All isotope decay follows the equation:

ln [A] = -kt + ln [A]₀

<em>Where [A] is actual amount of the isotope after time t, k is decay constant and [A]₀ the initial amount of the isotope</em>

We can find k from half-life as follows:

k = ln 2 / Half-Life

k = ln2 / 27.7 days

k = 0.025 days⁻¹

t = 48.2 days

[A]  = ?

[A]₀ = 39.7mg

ln [A] = -0.025 days⁻¹*48.2 days + ln [39.7mg]

ln[A] = 2.476

[A] = 11.9g remains after 48.2 days

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8 0
3 years ago
If this decay has a half life of 2.60 years, what mass of 72.5 g of sodium 22 will remain after 15.6 years
Vlad1618 [11]

Sodium-22 remain : 1.13 g

<h3>Further explanation </h3>

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.  

Usually, radioactive elements have an unstable atomic nucleus.  

General formulas used in decay:  

\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{T/t\frac{1}{2} }}}

T = duration of decay  

t 1/2 = half-life  

N₀ = the number of initial radioactive atoms  

Nt = the number of radioactive atoms left after decaying during T time  

half-life = t 1/2=2.6 years

T=15.6 years

No=72.5 g

\tt Nt=72.5.\dfrac{1}{2}^{15.6/2.6}\\\\Nt=72.5.\dfrac{1}{2}^6\\\\Nt=1.13~g

8 0
4 years ago
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