1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
insens350 [35]
3 years ago
13

DesperateDesperateDesperateDesperateDesperateDesperateDesperate

Chemistry
1 answer:
ale4655 [162]3 years ago
5 0
It moved up one unit because of the dinosaur
You might be interested in
What is Markanikov rule?<br>​
lana [24]
Markovnikov rule, in organic chemistry, a generalization, formulated by Vladimir Vasilyevich Markovnikov in 1869, stating that in addition reactions to unsymmetrical alkenes, the electron-rich component of the reagent adds to the carbon atom with fewer hydrogen atoms bonded to it, while the electron-deficient component ...
5 0
2 years ago
Read 2 more answers
What is the term for temperature in a room reaching the point where all exposed surfaces in room ignite?
eimsori [14]

Answer:flashover

Explanation: flashover are event that occurs when all of the combustible materials in a room reach their ignition temperatures at the same time

3 0
3 years ago
How many molecules are in 3.0 moles of water
diamong [38]
Hopefully this will help you my friend.

7 0
3 years ago
63. Use VSEPR theory to predict the structure of these
statuscvo [17]
Sulphate ion is used but fornyour kind information there ain't no place
4 0
3 years ago
Read 2 more answers
An unknown compound, X is thought to have a carboxyl group with a pKa of 2.0 and another ionizable group with a pKa between 5 an
Westkost [7]

Answer:

7.3

Explanation:

By Henderson Hasselbalch equation we can calculate the pH or the pOH of a solution by its pKa. Remember that pH = -log[H^{+}], and pKa = -logKa. Ka is the equilibrium constant of the acid.

Henderson Hasselbalch equation :

pH = pKa - log \frac{[HA]}{[A^{-}]}

Where [HA] is the concentration of the acid, and [A^{-}] is the concentration of the anion which forms the acid.

So, acid X, has two ionic forms, the carboxyl group and the other one. First, we have 0.1 mol/L of the acid, in 100 mL, so the number of moles of X

n1 = (0.1 mol/L)x(0.1 L) = 0.01 mol

When it dissociates, it forms 0.005 mol of the carboxyl group and 0.005 mol of the other group. Assuming same  stoichiometry.

Adding NaOH, with 0.1 mol/L and 75 mL, the number of moles of OH^- will be

n2 = (0.1 mol/L)x(0.075 L) = 0.0075 mol

So, the 0.0075 mol of OH^- reacts with 0.005 mol of carboxyl, remaining 0.0025 mol of OH^-, which will react with the 0.005 mol of the other group. So, it will remain 0.0025 mol of the other group.

The final volume of the solution will be 175 mL, but both concentrations (the acid form and ionic form) have the same volume, so we can use the number of mol in the equation.

Note that, the number of moles of the acid form is still 0.01 mol because it doesn't react!

So,

6.72 = pKa - log \frac{0.01}{0.0025}

6.72 = pKa - log 4

pKa - log4 = 6.72

pKa = 6.72 + log4

pKa = 6.72 + 0.6

pKa = 7.3

8 0
3 years ago
Other questions:
  • Describe the stages of gradual metamorphosis
    5·1 answer
  • If the net force on an object is 0 N, the forces are considered unbalanced. :true or false?​
    6·1 answer
  • What is happening in the East African Rift Zone?
    15·2 answers
  • ANDRES MEJIA-C...
    11·1 answer
  • 86.64 g LiBr mol LiBr
    8·2 answers
  • consider the reaction between acetylene C2H2 and oxygen in a welding torch 2C2H2 + 5O2 = 4CO2 + 2H2O of 5.4 moles of acetylene r
    14·1 answer
  • A chemical equation is balanced when the number of each type of __________ is the same on both sides of the yield sign.
    5·1 answer
  • I NEED THIS NOW AND NO LINKS OR ILL REPORT
    14·2 answers
  • Plz help me
    6·2 answers
  • The chemical equations in model 1 contain the phase notations (s), (l), (g), and (aq). match each symbol with its meaning.
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!