Which one of the following would form a precipitate when mixed with LiOH?

A) KNO3
B) NH4Cl
C) Ca(C2H3O2)2
D) ZnBr2

Answers

Answer 1

Answer:

ZnBr2

Explanation:

KNO3 + LiOH -------> no reaction

This is because KNO3 and LiOH completely ionize.in water and form neutral solutions.Since both are neutral no reaction occurs

NH4Cl + LiOH -----> NH3 (aq) + H2O (l) + LiCl(aq)

None of the above products are precipitates

Ca(C2H3O2)2 + 2LiOH -----> Ca(OH)2 (aq) + 2LiC2H3O2 (aq)

ZnBr2 + 2LiOH -----> Zn(OH)2 (s) + LiBr2 ( aq)

Zn(OH)2 thus formed is a white precipitate

However when excess LiOH is added Zn(OH)2 precipitate will dissolve to give a clear solution of Li2ZnO2 .

You can remember this by the fact that Na,K,Rb,Cs,Ca,Sr,Ba hydroxides are soluble in water and all other hydroxide are precipitated in water

Zn(OH)2 (s) + 2 LiOH ------> Li2ZnO2(aq) + 2 H2O ( l)

Answer 2

The only compound that will form a precipitate with LiOH would be ZnBr2

Precipitation reactions

These are double displacement reactions in which precipitates are produced. Precipitates are insoluble solids formed from the combination of ions in aqueous solutions.

In this case, a reaction between LiOH and ZnBr2 would be as follows;

2LiOH + ZnBr2 ----------> 2LiBr + Zn(OH)2 (s)

Zn(OH)2 is an insoluble salt and will precipitate out in the solution.

More on precipitation can be found here: https://brainly.com/question/5019199


Related Questions

I NEED THE ANSWER ASAP! Its science btw

Which statement describes conditions in which a mineral can form?
A. Crystals of minerals dissolve in the groundwater in caves.

B. Materials dissolved in seawater crystallize on an ocean bottom.

C. Molten materials are cooled in a netalworks factory

D. Materials are mined from deposits deep underground.​

Answers

A. Crystals of minerals dissolve in the groundwater in caves

Answer:

b

Explanation:

Solid calcium hydroxide is dissolved in water until the pH of the solution is 11.44. The hydroxide ion concentration [OH–] of the solution is:______.A. 1.1 * 10-11 M.B. 3.06 M.C. 8.7 * 10-4 M.D. 1.0 * 10-14 M.E. None of these.

Answers

Answer:

c the answer is c that is the answer

You have a sealed 2 liter flask that contains nothing but water and carbon dioxide. The flask is half-filled with liquid water, has a temperature of 25°c, and the overall pressure within the flask is 0. 1 atm. How many moles of co2 are in the flask? at this temperature, you may take the kh value for co2 as 0. 033 m / atm.

Answers

In this exercise we want to calculate the amount of moles, so this is going to be:

[tex](4.6)(10^{-3}) \ mols \ CO_2[/tex]

Knowing that Henry's law is given by:

[tex]C = KHP[/tex]

Where constants are given by:

C = Concentration  KH = Henry's law constant = [tex]0.033 m/atm[/tex]P = partial pressure = [tex]0.07 atm[/tex]

Before we can find the concentration of CO2 (and hence the moles of CO2), we first need to find its partial pressure.  We look up the vapor pressure of water at 25º and find it to be 0.03 atm.  Since the total pressure is equal to 0.1 atm, this mean the partial pressure of:

 

[tex]CO_2 = 0.1 \ atm - 0.03 \ atm = 0.07 \ atm[/tex]

Now using Henry's law, we find the concentration:

[tex]C = (0.033)*( 0.07) = (2.31)*(10^{-3})[/tex]

Converting to moles of CO2, we have:

[tex](2.31)*(10^{-3})*( 2) = (4.6)*(10^{-3})[/tex]

See more about concentration at https://brainly.com/question/3045247

22) Calculate the energy of a photon of radiation with a frequency of 9.50 x 1013 Hz.

Answers

The energy of the photon with a frequency of 9.50×10¹³ Hz is 6.29×10¯²⁰ J

From the question given above, the following data were obtained:

Frequency (f) = 9.50×10¹³ Hz

Planck's constant (h) = 6.626×10¯³⁴ Js

Energy (E) =?

The energy of the photon can be obtained as follow:

E = hf

E = 6.626×10¯³⁴ × 9.50×10¹³

E = 6.29×10¯²⁰ J

Therefore, the energy of the photon is 6.29×10¯²⁰ J

Learn more: https://brainly.com/question/7957705

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