Ask question

Ask question

All categories

3 years ago

9

Calculate the pH in the titration of 25.0 mL of 0.100 M acetic acid by sodium hydroxide after the addition to the acid solution

of 10.0 mL of 0.100 M NaOH. (Ka for acetic acid is 1.8x10-5.)A.8.72B. 12.22C. 7.14D. 4.57

1 answer:

Marysya12 [62]3 years ago

6 0

Answer:

see the pic for more detail

You might be interested in

Wch of the following phrases is the best definition of science?

krek1111 [17]

Answer: A i think

Explanation:

7 0

3 years ago

Read 2 more answers

The question is below!

Agata [3.3K]

The anode is the negative electrode and so will be donating electrons to assist in this chemical reaction occuring. All reactions accept electrons as reactants. The key issue is the reduction potential Eo (+1.8V). This is greatest for the reaction:

Co3+ + e -> Co2+

Therefore this reaction has the greatest tendency to occur.

6 0

3 years ago

The maximum amount of nickel(II) cyanide that will dissolve in a 0.220 M nickel(II) nitrate solution is...?

sweet [91]

Answer : The maximum amount of nickel(II) cyanide is $5.84\times 10^{-12}M$

Explanation :

The solubility equilibrium reaction will be:

$Ni(CN)_2\rightleftharpoons Ni^{2+}+2CN^-$

Initial conc. 0.220 0

At eqm. (0.220+s) 2s

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ni^{2+}][CN^-]^2$

Now put all the given values in this expression, we get:

$3.0\times 10^{-23}=(0.220+s)\times (2s)^2$

$s=5.84\times 10^{-12}M$

Therefore, the maximum amount of nickel(II) cyanide is $5.84\times 10^{-12}M$

7 0

3 years ago

You are given a cube of pure copper when you calculate the density using

lara [203]

Answer:

2.01% to the nearest hundredth.

Explanation:

Percent error =[ (8.96-8.78) / 8.96]* 100

= 0.020089 * 100

= 2.0089 %

5 0

3 years ago

Sodium electron configuration

Alex73 [517]

Full:

1s² 2s² 2p⁶ 3s¹

Abbreviated:

[Ne] 3s¹

4 0

3 years ago