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klio [65]
3 years ago
10

Calculate the pH of a solution with an H+ ion concentration of 9.36 x 10-3 M ?

Chemistry
1 answer:
mel-nik [20]3 years ago
6 0

Answer:

The answer is 2.03

Explanation:

The pH of a solution can be found by using the formula

pH = - log [ {H}^{+} ]

where H+ is the Hydrogen ion concentration

From the question we have

pH =  -  log(9.36 \times  {10}^{ - 3} )  \\  = 2.02872415...

We have the final answer as

<h3>2.03 </h3>

Hope this helps you

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What happend if NH3+ HJ
Liula [17]
Are you sure you copied that down correctly? I know what NH3 is but HJ I am comming up blank on. The only thing HJ I know is hebdojoule which is unit system.
Are you doing chemical balancing?
6 0
3 years ago
Which element has the greatest density at stp? scandium, selenium, silicon, or sodium?
Slav-nsk [51]
The answer is selenium
4 0
3 years ago
What is the definition of an Arrhenius acid?
Lorico [155]

Answer:

I <u>think</u> your answer is: C. ("An Arrhenius acid increases [H +] in the solution.")

Explanation:

An Arrhenius acid is a substance that dissociates in water to form hydrogen ions (H + ); that is, an acid increases the concentration of H + ions in an aqueous solution. This causes the protonation of water, or the creation of the hydronium (H 3 O +) ion.

Hopefully this helps!

Have a great day! ^^

5 0
3 years ago
How many moles of ions are in 1.45 mol of K2SO4 ?
leonid [27]

Answer:

Explanation:

First we look generally at what makes K2SO4.

In one mole of K2SO4, there are 2 moles of the potassium ion (K+) and 1 mole of sulfate ion (SO4 2-).

Knowing that; in 1.75 moles of K2SO4, there must be 2 x 1.75moles of potassium ion (K+) and 1 x 1.75moles of Sulfate ion (SO4 2-)

This gives us 3.5moles of K+ and 1.75moles of SO4 2-

3 0
3 years ago
A thin-walled sphere rolls along the floor. What is the ratio of its translational kinetic energy to its rotational kinetic ener
ICE Princess25 [194]

Explanation:

The kinetic energy of translation

E_1=\frac{1}{2}mv^2

m= mass v= linear velocity

The kinetic energy of rotation

E_2=\frac{1}{2}I\omega^2

I= MOI of the thin walled sphere =kmR^2

where ω= v/R= angular velocity

E_2=\frac{1}{2}kmR^2\frac{v}{R}^2

Then

\frac{E_1}{E_2} = \frac{0.5mv^2}{0.5kmR^2\frac{v}{R}^2 }

=1/k

solid sphere: k=0.4;   E1/E2 =1/0.4 = 2.5;  

 hollow sphere: k=2/3;   E1/E2 = 1.5

3 0
3 years ago
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