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What is the concentration of H^ + at apH = 2 ? Mol / L What is the concentration of H^ + ions at apH = 6 ? Mol/L How many more H

Nonamiya [84]

Answer:

The concentration of hydrogen ion at pH is equal to 2 : $= [H^+]=0.01 mol/L$

The concentration of hydrogen ion at pH is equal to 6 : $[H^+]'=0.000001 mol/L$

There are 0.009999 more moles of $H^+$ ions in a solution at a pH = 2 than in a solution at a pH = 6.

Explanation:

The pH of the solution is the negative logarithm of hydrogen ion concentration in an aqueous solution.

$pH=-\log [H^+]$

The hydrogen ion concentration at pH is equal to 2 = [H^+]

$2=-\log [H^+]\\$

$[H^+]=10^{-2}M= 0.01 M=0.01 mol/L$

The hydrogen ion concentration at pH is equal to 6 = [H^+]

$6=-\log [H^+]\\\\$

$[H^+]=10^{-6}M= 0.000001 M= 0.000001 mol/L$

Concentration of hydrogen ion at pH is equal to 2 = $[H^+]=0.01 mol/L$

Concentration of hydrogen ion at pH is equal to 6 = $[H^+]'=0.000001 mol/L$

The difference between hydrogen ion concentration at pH 2 and pH 6 :

$= [H^+]-[H^+]' = 0.01 mol/L- 0.000001 mol/L = 0.009999 mol/L$

Moles of hydrogen ion in 0.009999 mol/L solution :

$=0.009999 mol/L\times 1 L=0.009999 mol$

There are 0.009999 more moles of $H^+$ ions in a solution at a pH = 2 than in a solution at a pH = 6.

8 0

3 years ago

the refractive index is the blank of the speed of light in a medium to the speed of light in a vacuum

andrew11 [14]

Answer:ratio

Explanation:

8 0

3 years ago

Read 2 more answers

g stAn experienced spear fisherman sees a small fish swimming in a tidal pool. If the fisherman sees the fish at approximately a

Luba_88 [7]

Answer:

θ = 28.9

Explanation:

For this exercise let's use the law of refraction

n₁ sin θ₁ = n₂ sin θ₂

where we use index 1 for air and index 2 for water where the fish is

sin θ₂ = n₁ / n₂ sin θ₁

in this case the air repair index is 1 and the water 1.33

we substitute

sin θ₂ = 1 / 1.33 sin t 40

sin θ = 0.4833

θ = sin⁻¹ 0.4833

θ = 28.9

5 0

3 years ago

Let's say you go on a run and start to sweat because you are heating up.

Jobisdone [24]

Answer:

kinetic to thermal

Explanation:

4 0

3 years ago

I wish to use a step up transformer to turn an initial RMS AC voltage of 100 V into a final RMS AC voltage of 200 V. What is the

zhuklara [117]

Answer:

1:2

Explanation:

It is given that,

Initial RMS AC voltage is 100 V and final RMS AC voltage is 200 V.

We need to find the ratio of the number of turns in the primary to the secondary for step up transformer.

For a transformer, $\dfrac{V_1}{V_2}=\dfrac{N_1}{N_2}$

So,

$\dfrac{N_1}{N_2}=\dfrac{100}{200}\\\\\dfrac{N_1}{N_2}=\dfrac{1}{2}$

So, the ratio of the number of turns in the primary to the secondary is 1:2.

4 0

3 years ago