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statuscvo [17]
3 years ago
6

You are performing a titration by adding a Strong Base to a solution of Weak Acid. At the equivalence point, will your solution

be acidic, basic, or neutral?
Chemistry
1 answer:
Alchen [17]3 years ago
3 0

Answer:

Basic

Explanation:

The titration of a strong base with a weak acid would always result in a solution whose pH is greater than 7,i.e a basic solution. For this reason its advisable to use an indicator whose color change will occur in that pH range.

For other cases, the titration of a stong acid and strong will result in a neutral solution and the titration of a weak base versus a strong acid would result in an acidic solution.

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Hence they got seven electrons in ground state
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HURRY FOR A TEST! The ionization energies for removing successive electrons from sodium are 496 kJ/mol, 4562 kJ/mol,
prohojiy [21]

Answer:

D. The noble gas configuration has been reached.

Explanation:

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3 years ago
Initially, 0.65 mol of PCl5 is placed in a 1.0 L flask. At equilibrium, there is 0.15 mol of PCl3 in the flask. What is the equi
borishaifa [10]

Answer : The equilibrium concentration of PCl_5 is, 0.50 M

Explanation : Given,

Initial moles of PCl_5 = 0.65 mole

Volume of solution = 1.0 L

Moles of PCl_3 at equilibrium = 0.15 mole

The balanced equilibrium reaction will be,

                          PCl_5\rightleftharpoons PCl_3+Cl_2

Initial moles     0.65        0         0

At eqm.           (0.65-x)     x         x

Moles of PCl_3 at equilibrium = x = 0.15 mole

Moles of Cl_2 at equilibrium = x = 0.15 mole

Moles of PCl_5 at equilibrium = (0.65-x) = (0.65-0.15) = 0.50 mole

Now we have to calculate the concentration of PCl_3,Cl_2\text{ and }PCl_5 at equilibrium.

Formula used : Concentration=\frac{Moles}{Volume}

\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}=\frac{0.15mole}{1.0L}=0.15M

\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}=\frac{0.15mole}{1.0L}=0.15M

\text{Concentration of }PCl_5=\frac{\text{Moles of }PCl_5}{\text{Volume of solution}}=\frac{0.50mole}{1.0L}=0.50M

Therefore, the equilibrium concentration of PCl_5 is, 0.50 M

8 0
4 years ago
All of the food that we eat, liquids that we drink and medications that we take are
kirill115 [55]

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3 0
4 years ago
A solution of rubbing alcohol is 75.7 % (v/v) isopropanol in water. How many milliliters of isopropanol are in a 87.7 mL sample
beks73 [17]

Answer:

66.4 mL

Explanation:

A 75.7% (v/v) value, means that f<u>or every 100 mL of rubbing alcohol, there are 75.7 mL of isopropanol.</u>

With the above information in mind, we can s<u>olve the problem by multiplying 87.7 mL by 75.7 %</u>:

87.7 mL * 75.7 / 100 = 66.4 mL

So there are 66.4 mL of isopropanol in 88.7 mL of rubbing alcohol.

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3 years ago
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