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Rama09 [41]
3 years ago
7

What is the concentration of H+ ions at a pH = 11?

Chemistry
1 answer:
natta225 [31]3 years ago
6 0

Answer:

\huge 1 × {10}^{-11} \: \: M

Explanation:

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

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

Since we are finding the H+ ions we find the antilog of the pH

So we have

11 =  -  log({H}^{+})  \\ {H}^{+} =  {10}^{ - 11}

We have the final answer as

1 × {10}^{-11} \: \: M

Hope this helps you

You might be interested in
How does the appearance of a substance change when it<br> changes phase?
kicyunya [14]

Answer:

Melting: the substance changes back from the solid to the liquid. Condensation: the substance changes from a gas to a liquid. Vaporization: the substance changes from a liquid to a gas. Sublimation: the substance changes directly from a solid to a gas without going through the liquid phase.

Explanation:

6 0
3 years ago
A 1.0 mole sample of fluorine gas at 25 °C has an average molecular velocity of 415 m/s. What is the total KE of the gas sample?
Mariana [72]

The total kinetic energy of the gas sample is 3.3 KJ

<h3>What is kinetic energy? </h3>

This is the energy possessed by an object in motion. Mathematically, it can be expressed as:

KE = ½mv²

Where

  • KE is the kinetic energy
  • m is the mass
  • v is the velocity

<h3>How to determine the mass of the fluorine gas</h3>
  • Molar mass of fluorine gas = 38 g/mol
  • Mole of fluorine gas = 1 mole
  • Mass of fluorine gas = ?

Mass = mole × molar mass

Mass of fluorine gas = 1 × 38

Mass of fluorine gas = 38 g

<h3>How to determine the KE of the gas sample</h3>
  • Mass (m) = 38 g = 38 / 1000 = 0.038 Kg
  • Velocity (v) = 415 m/s
  • Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 0.038 × 415²

KE = 3272.275 J

Divide by 1000 to express in kilojoule

KE = 3272.275 / 1000

KE = 3.3 KJ

Learn more about energy:

brainly.com/question/10703928

#SPJ1

4 0
2 years ago
A 35.0-ml sample of 0.20 m lioh is titrated with 0.25 m hcl. What is the ph of the solution after 23.0 ml of hcl have been added
kiruha [24]
<h3><u>Answer;</u></h3>

pH = 12.33

<h3><u>Explanation;</u></h3>

The equation of reaction is :

LiOH(aq) + HCl(aq) --> LiCl(aq) + H2O(l)

Reactants left after the titrant is added;

Total Moles LiOH;

= 0.035L LiOH × (0.2moles/L)

= 0.007moles of LiOH

Moles of HCl;

= 0.023L HCl × (0.25moles/L)

= 0.00575moles HCl is the limiting reagent

Reacting amount of moles of LiOH;

= 0.0575 moles HCl *(1mole LiOH/1moles HCl)

=0.00575 moles LiOH (reacted)

Moles of LiOH left;

= 0.007moles total - 0.00575moles that react

= .00125 moles of LiOH (left)

LiOH is a strong base, which means that it ionizes completely.  

0.00125moles LiOH *(moles/0.058L) = 0.02155M of LiOH

LiOH(aq) --> Li+(aq) + OH-(aq)

[LiOH] = [OH-] = 0.02155 M

pOH = -log[OH-]

pOH = -log(0.02155)

pOH= 1.67

pH = 14 - pOH

pH = 14 - 1.67

pH = 12.33

7 0
3 years ago
Write the structure of the product that would be formed from the S(
prisoha [69]

Answer:The product formed on reaction with hydroxide ion as nucleophile is 2R-hexane-2-ol.

The product formed on reaction with water would be a 50:50 mixture of

2S-hexane-2-ol. and 2R-hexane-2-ol.

Explanation:

2S-iodohexane on reactiong with hydroxide ion would undergo SN² substitution reaction that is substitution bimolecular. Hydroxide ion has a negative charge and hence it is a quite good  nucleophile .

The rate of a SN² reaction depends on both the substrate and nucleophile . Here the substrate is a secondary carbon center having Iodine as a leaving group.SN² reaction  takes place here as hydroxide ion is a good nucleophile and it can attack the secondary carbon center from the back side leading to the formation of 2R-hexane-2-ol.

In a SN² reaction since the the nucleophile attacks from the back-side so the product formation takes place with the inversion of configuration.

When the same substrate S-2-iodohexane undergoes a substitution reaction with water as a nucleophile then the reaction occurs through (SN¹) substitution nucleophilic unimolecular mechanism .

The rate of a SN¹ reaction depends only on the nature of substrate and is independent of the nature of nucleophile.

The SN¹ reaction is a 2 step reaction , in  the first step leaving group leaves leading to the formation of a carbocation and once the carbocation is formed then any weaker nucleophile or even solvent molecules can attack leading the formation of products.

In this case a secondary carbocation would be generated in the first step and then water will attack this carbocation to form the product in the second step.

The product formed on using water as a nucleophile would be a racemic mixture of R and S isomers of hexane -2-ol in 50:50 ratio. The two products formed would be 2R-hexane-2-ol and 2S-hexane-2-ol.

Kindly refer the attachment for reaction mechanism and structure of products.

8 0
3 years ago
PLZZ HELP ASAPPPJKSOWLSOWOLWKSKW
Minchanka [31]
I’m pretty the answer would be continental slope. :)
I really hope this helps.
4 0
3 years ago
Read 2 more answers
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