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allochka39001 [22]

2 years ago

10

The pH level of a solution measures how acidic or basic the solution is. Values between 0 and 7 are considered acidic while valu

es between 7 and 14 are considered basic. The pH is calculated by using the equation pH=−log[H3O+]. The notation [H3O+] represents the concentration of hydronium ions, which is measured in units of moles per liter, M.
What is the pH level of a solution if [H3O+]=3.4×10^−3M?

1 answer:

Furkat [3]2 years ago

4 0

Answer:

about 2.47

Step-by-step explanation:

Your calculator will tell you ...

$-\log{(3.4\times 10^{-3})}\approx 2.4685\approx\bf{2.47}$

You might be interested in

Solve for z<br> 4/z =<br> 12/5 <br> what is z

Verdich [7]

Answer:

5/3 = z

Step-by-step explanation:

You can use proportions to figure this out.

Imma solve this as I work through it.

So

4/z is equal to 12/5. So you can do something called the butterfly method (I have never used this before, but it works tho) and the method involves only multiplication and division.

Lemme clean this up for ya

The butterfly method involves multiplying in a cross format on both sides, so the

4 12

__ = ______

z 5

becomes 4 times 5, and z times 12.

4 times 5 is 20, and z times 12 is 12z

so you now have 20 = 12z

Lets simplify the equation by dividing by 4s

20 divided by 4 is 5 and 12z divided by 4 is 3z

so

5 = 3z

You can divide again

by dividing both sides

by

3

to get the unit rate of z.

It will be a fraction though

and it won't look pretty

but here it is:

5

__ = z

3

So

It is

5/3

Hope this helps

and hope you get a laugh out of this

yeeha

7 0

2 years ago

Both the La Plata river dolphin (Pontoporia blainvillei) and

Citrus2011 [14]

Answer:

<em>1. A sperm whale is </em><em>3</em><em> </em><em>orders of magnitude</em><em> heavier than a La Plata river dolphin; 2. The radius of the larger ball is </em><em>one (1) order of magnitude</em><em> bigger than the radius of the smaller ball; 3. The volume of the larger ball is </em><em>3 orders of magnitude</em><em> bigger than the volume of the smaller ball</em>.

Step-by-step explanation:

If we expressed a number as:

$\\ N = a * 10^{b}$ (1)

Where

$\\ \frac{1}{\sqrt{10}} \leq a < \sqrt{10}$ (2)

or

$\\ 1 \leq a < 10$ (3)

Then, <em>b</em> represents the <em>order of magnitude </em>of such a number (<em>Order of magnitude (2020), </em>in Wikipedia).

The order of magnitude can be defined as "...the smallest power of ten needed to represent a quantity" (Weisstein, Eric W. "Order of Magnitude". From MathWorld--A Wolfram Web Resource).

Having gathered all this information, we can proceed as follows:

<h3>First case</h3>

The<em> La Plata river dolphin</em> weighs between 30 and 50kg and the sperm whale weighs between 35,000 and 40,000kg.

Then, considering (1) and (3) to express the dolphin and whale's weight (since in this way the order of magnitude is the same as the exponent part in the <em>scientific notation</em>):

$\\ 30kg \leq Dolphin_{weight} \leq 50kg$

$\\ 3*10^{1}kg \leq Dolphin_{weight} \leq 5*10^{1}kg$

$\\ 35000kg \leq Whale_{weight} \leq 40000kg$

$\\ 3.5*10^{4}kg \leq Whale_{weight} \leq 4.0*10^{4}kg$

Since the range for the weights are in the same order of magnitude for both dolphin and whale (considering the definition above):

$\\ Dolphin_{weight} = 10^{1}\;(order\;of\;magnitude=1)$

$\\ Whale_{weight} = 10^{4}\;(order\;of\;magnitude=4)$

Then

$\\ \frac{Whale_{weight} = 10^{4}}{Dolphin_{weight} = 10^{1}}$

$\\ \frac{Whale_{weight}}{Dolphin_{weight}} = \frac{10^{4}}{10^{1}}$

$\\ \frac{Whale_{weight}}{Dolphin_{weight}} = 10^{4-1}$

$\\ \frac{Whale_{weight}}{Dolphin_{weight}} = 10^{3}$

Thus

<em>A sperm whale is </em><em>3</em><em> </em><em>orders of magnitude</em><em> heavier than a La Plata river dolphin.</em>

<h3>Second case</h3>

Following the same reasoning, we can conclude that <em>the radius of the larger ball is </em><em>one (1) order of magnitude</em><em> bigger than the radius of the smaller ball:</em>

$\\ \frac{Larger\;ball_{radius}}{Smaller\;ball_{radius}} = \frac{10^{1}}{10^{0}}$

$\\ \frac{Larger\;ball_{radius}}{Smaller\;ball_{radius}} = 10^{1-0} = 10^{1}$

<h3>Third case</h3>

For this case, we need to calculate <em>the volume of a sphere</em> for both radii (1cm and 10cm).

The volume of a sphere is

$\\ V_{sphere} = \frac{4}{3}*\pi*R^{3}$

Then, the volume of the <em>ball of radius 1cm</em> is:

$\\ V_{radius=1} = \frac{4}{3}*\pi*(1cm)^{3}$

$\\ V_{radius=1} \approx 4.19*10^{0}cm^{3}$

And, the volume of the <em>ball of radius 10cm</em> is:

$\\ V_{radius=10} = \frac{4}{3}*\pi*(10cm)^{3}$

$\\ V_{radius=10} \approx 4.19*10^{3}cm^{3}$

Thus

$\\ \frac{10^{3}}{10^{0}} = 10^{3}$

As a result, <em>the volume of the larger ball is </em><em>3 orders of magnitude</em><em> bigger than the volume of the smaller ball</em>.

4 0

3 years ago

Is -1.125 a real number?

pochemuha

$-1.125=-1\frac{125}{1000}=-1\frac{1}{8}\ is\ real\ number$

7 0

3 years ago

Read 2 more answers

The line plot shows the number of different colors that art students used in a colored-pencil sketch. How many art students used

Blababa [14]

Where is the line plot?

4 0

2 years ago

Two fifth of a certain number is 30 what is the number?

vesna_86 [32]

<h2>75</h2>

- - - - - - - - - - - - - - - - - - - - - - - -

Beginning with an "If-then" statement ⇣

<em>If </em><em>2/5 = 30 </em><em>then</em><em> 1/5 = 15</em>

Now we have 1/5, let's make it 1 by multiplying x5.

<em>15 x 5 = </em><em>75</em>

- - - - - - - - - - - - - - - - - - - - - - - -

Good luck! :)

~pinetree

3 0

2 years ago