Please help on how to solve this problem it’s for isotopes

what is the molarity of 20.0 ml of a KCl solution that reacts completely with 30.0 ml of a 0.400 M Pb(NO3)2 solution?

Alex17521 [72]

Answer:

[KCl] = 1.2 M

Explanation:

We need to complete the reaction:

2KCl(aq) + Pb(NO₃)₂(aq) → 2KNO₃(aq) + PbCl₂(s)↓

By stoichiomety we know that 1 mol of chloride needs 1 mol of nitrate to react:

Let's find out the moles of nitrate, we have:

Molarity = mol/volume(L)

We convert the volume → 30 mL . 1L/1000mL = 0.030L

Molarity . volume(L) = moles → 0.400 M . 0.030L = 0.012 moles

Therefore, we can make a rule of three.

1 mol of nitrate reacts with 2 moles of chloride

Then, 0.012 moles of nitrate must react with (0.012 . 2) / 1 = 0.024 moles of KCl

We convert the volume from mL to L → 20 mL . 1L /1000mL = 0.020L

Molarity = mol /volume(L) → 0.024 mol /0.020L = 1.2 M

7 0

3 years ago

The NH₄⁺ ion forms acidic solutions, and the CH₃COO⁻ ion forms basic solutions. However, a solution of ammonium acetate is almos

tigry1 [53]

Not always ammonium salts of weak acids form neutral solutions.

When formic acid reacts with ammonia, ammonium formate is produced:

HCO2H + NH3 ----> NH4HCO2

You already know that the weak conjugate bases of NH3 and HCO2H are NH4+ and HCO2, respectively.

How can the pH of the solution be calculated if the salt's anion causes the pH to rise and the salt's cation causes it to fall? The relative intensities of the basic anion and the acidic cation hold the key to the solution.

As was already established, formate is a weak base and will create hydroxide ions in water, whereas ammonium is a weak acid and will make hydronium ions in water.

NH4⁺ + H2O -----> NH3 + H3O⁺

HCO2⁻ + H2O -----> HCO2H + OH⁻

Since the acid ionization of NH4+ is more favored than the base ionization of HCO2-, the solution will be acidic.

To learn more about ammonium salts:

brainly.com/question/10874844

#SPJ4

6 0

1 year ago

In a 1.0x10^-4 M solution of HClO(aq), identify the relative molar amounts of these species:HClO, OH-, H3O+, OCl-, H2O

yarga [219]

HClO is a weak acid, which means the ions do not fully dissociate. The hydrolysis reaction for the hypochlorous acid is:

HClO + H2O ⇄ H3O+ +OCl-

Then the equilibrium constant, Ka, of dilute HClO would be:

$K_{a} = \frac{[ H_{3} O^{+} ][O Cl^{-} ]}{HClO}$

Then we do the ICE table. I is for the initial concentration, C for the change and E for the excess.

HClO + H2O ⇄ H3O+ + OCl-
I 1.0x10^-4 0 0
C -x +x +x
E (1.0x10^-4 - x) x x

Substituting the excess (E) concentration to the Ka equation:

$K_{a} = \frac{[x ][x]}{1.0 \ x \ 10^{-4} - x }$

Simplifying the equation would yield a quadratic equation:

$x^{2} + K_{a}x-(1.0 \ x \ 10^{-4}) K_{a}=0$

The Ka for HClO is an experimental data which was determined to be 2.9 x 10^-8. Substitute this to the equation, determine the roots, then you get the value for x, which is the concentration of H3O+ and ClO-. Just use your calculator feature Shift-Solve.

x = 1.688 x 10^-6 M = [H3O+] = [ClO-]

Then, you can determine the conc of [OH-] through pH.

pH = -log {H3O+] = -log [1.688 x 10^-6] = 5.77
pOH = 14 - pH = 14 - 5.77 = 8.23
pOH = 8.23 = -log [OH-]
[OH-] = 5.89 x 10^-9 M

Also, since HClO is (1.0x10^-4 - x), then it's concentration would be:
[HClO] = 1.0x10^-4 - 1.688 x 10^-6 = 9.83 x10^-5 M

Let's summarize all concentrations:
[HClO] = 9.83 x10^-5 M
[OH-] = 5.89 x 10^-9 M
[H3O+] = [ClO-] = 1.688 x 10^-6 M
Since the solution is dilute, H2O is relatively higher in concentration.

Thus in relative amounts, the order would be

H2O >>> HClO > H3O+ = ClO- > OH-

6 0

3 years ago

Read 2 more answers

4. Quando se obtém a potência máxima da célula, o valor indicado no amperímetro, na unidade SI, está no intervalo: (A) [3,1; 3,2

Annette [7]

Answer:i dont speak english

Explanation:

sorry

8 0

3 years ago

HELP 100 PONTS GIVING BRAINLIEST

Oksana_A [137]