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3 years ago

If the hydrogen ion concentration [H+] of an aqueous solution is 1 x 10-9, what is the pH?

Mathematics

1 answer:

RSB [31]3 years ago

3 0

Answer:

Step-by-step explanation: In this way, pH is determined by hydrogen-ion concentration. In the case of a neutral solution, [H+]=10-7 , which we call a pH of 7. This means, for example, that a hydrogen-ion concentration of a solution with a pH of 4 is 10-4mol/l, meaning it contains 0.0001 mol of hydrogen ions in a solution of 1 liter.

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Work out Ryan’s average speed for his total return journey detailing how you worked it out.

pickupchik [31]

Answer:

For the return, i guess that you say in the interval between 15:40 and 15:45, that is the moment where he goes back to his home, and pass through the local shop.

Ok, in the graph we can see pairs (time, position)

The average speed can be obtained with two of these pairs, and the average speed is:

As = (position2 - position1)/(time2 - time1)

In this particular case, we have the pairs:

(15:40, 900m) and (15:45, 400m)

(i guess that the distance is in meters)

Then the average speed is:

As = (900m - 400m)/(15:45 - 15:40)

and 15:45 - 15:40 = 0.5 minutes.

As= 500m/5 min = 100 m/min

This means that the average speed in this return is 100 meters per minute

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3 years ago

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qaws [65]

Answer:sheep

Step-by-step explanation:

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3 years ago

What is 4.26 divided by 3

wlad13 [49]

The answer you are looking for is 1.42

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4 years ago

Read 2 more answers

Evaluate the sum of the following finite geometric series.

rjkz [21]

Answer:

$\large\boxed{\dfrac{156}{125}\approx1.2}$

Step-by-step explanation:

<h3>Method 1:</h3>

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\\\\for\ n=1\\\\\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1\\\\for\ n=2\\\\\left(\dfrac{1}{5}\right)^{2-1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}\\\\for\ n=3\\\\\left(\dfrac{1}{5}\right)^{3-1}=\left(\dfrac{1}{5}\right)^2=\dfrac{1}{25}\\\\for\ n=4\\\\\left(\dfrac{1}{5}\right)^{4-1}=\left(\dfrac{1}{5}\right)^3=\dfrac{1}{125}$

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}=\dfrac{125}{125}+\dfrac{25}{125}+\dfrac{5}{125}+\dfrac{1}{125}=\dfrac{156}{125}$

<h3>Method 2:</h3>

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\to a_n=\left(\dfrac{1}{5}\right)^{n-1}\\\\\text{The formula of a sum of terms of a geometric series:}\\\\S_n=a_1\cdot\dfrac{1-r^n}{1-r}\\\\r-\text{common ratio}\to r=\dfrac{a_{n+1}}{a_n}\\\\a_{n+1}=\left(\dfrac{1}{5}\right)^{n+1-1}=\left(\dfrac{1}{5}\right)^n\\\\r=\dfrac{\left(\frac{1}{5}\right)^n}{\left(\frac{1}{5}\right)^{n-1}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\r=\left(\dfrac{1}{5}\right)^{n-(n-1)}=\left(\dfrac{1}{5}\right)^{n-n+1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}$

$a_1=\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1$

$\text{Substitute}\ a_1=1,\ n=4,\ r=\dfrac{1}{5}:\\\\S_4=1\cdot\dfrac{1-\left(\frac{1}{5}\right)^4}{1-\frac{1}{5}}=\dfrac{1-\frac{1}{625}}{\frac{4}{5}}=\dfrac{624}{625}\cdot\dfrac{5}{4}=\dfrac{156}{125}$

5 0

3 years ago

Solve the system by substitution.<br> y = -3x +4<br> y = x

mafiozo [28]

Answer:

Step-by-step explanation:

7 0

3 years ago