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4 years ago

For which salt should the aqueous solubility be most sensitive to ph?

1 answer:

6 0

The salt that its aqueous solubility most sensitive to pH would be calcium fluoride (CaF2). It is an insoluble salt which when reacted with an acid, it yields this equation:

CaF2 (s) + 2 HX (aq) → 2 HF (aq) + CaX2 (aq)

Since it produces H+ and and OH- when combined with water, it is therefore sensitive to pH.

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which statement regarding the relative energies of monochromatic light with with λ = 800 nm andmonochromatic light with λ = 400

olga2289 [7]

Answer:

The correct answer is option A.

Explanation:

Energy of the photon of an electromagnetic radiation is give by Planck's equation:

$E=\frac{hc}{\lambda }$

Where:

h = Planck's constant

c = speed of light

$\lambda$ = Wavelength of the photon's radiation

Energy of photon of 800 nm monochromatic light

$\lambda = 800 nm=400\times 10^{-9} m$

$E=\frac{hc}{800 \times 10^{-9} m}$

Energy per mole of photons:

$E\times N_A=\frac{hc}{800 \times 10^{-9} m}\times 6.022\times 10^{23} mol^{-1}$ ..[1]

Energy of photon of 400 nm monochromatic light

$\lambda '= 400 nm=400\times 10^{-9} m$

Energy per mole of photons:

$E'\times N_A=\frac{hc}{400 \times 10^{-9} m}\times 6.022\times 10^{23} mol^{-1}$ ..[2]

[1] ÷ [2]

$\frac{E\times N_A}{E'\times N_A}=\frac{\frac{hc\trimes 6.022\times 10^{23} mol^{-1}}{800 \times 10^{-9} m}}{\frac{hc\times 6.022\times 10^{23} mol^{-1}}{400 \times 10^{-9} m}}$

$\frac{E}{E'}=\frac{1}{2}$

$E=\frac{1}{2}\times E'$

8 0

3 years ago

Read 2 more answers

Convert 22.5 g S to moles of S.

Neporo4naja [7]

Grams of compound
moles——————————————-
Molar mass of compound

moles=22.5g S
-————
32.06 g/mol

moles=0.701

8 0

4 years ago

Phosphorus-32 has a half-life of 14.0 days. Starting with 6.00 g of 32P, how many grams will remain after 42.0 days ?

aleksandrvk [35]

6g...........3g..........1,5g........0,75g
0d..........14d.........28d.........42d

After 42 days 0,75g of Phoshorus-32 will remain.

4 0

3 years ago

Read 2 more answers

PlEASE HELP! 40!

podryga [215]

Answer:

moles Na = 0.1114 g / 22.9898 g/mol=0.004846

moles Tc = 0.4562g /98.9063 g/mol=0.004612

mass O = 0.8961 - ( 0.1114 + 0.4562)=03285 g

moles O = 0.3285 g/ 15.999 g/mol=0.02053

divide by the smallest

0.02053/ 0.004612 =4.45 => O

0.004846/ 0.004612 = 1.0 => Tc

to get whole numbers multiply by 2

Na2Tc2O 9

Explanation:

Hope it right hope it helps

5 0

3 years ago

a) Calculatethe molality, m, of an aqueous solution of 1.22 M sucrose, C12H22O11. The density of the solution is 1.12 g/mL.b) Wh

Contact [7]

Answer:

a) 1,74 molal

b) 37,2 %

c) 0,03

Explanation:

We are going to define sucrose as solute, water as solvent and the mix of both, the solution.

Let´s start with the data:

$Molarity = M = \frac{1,22 mol solute}{lts solution}$

We can assume as a calculus base, 1 liter of solution. So, in 1 liter of solution we have 1,22 moles of solute:

$1 lts solution * \frac{1,22 moles solute}{lts solution}=1,22 moles solute$

Knowing that the molality (m) is defined as mol of solute/kgs solvent, we have to calculate the mass of solvent on the solution. Remember our calculus base (1 lts of solution). In 1 lts of solution we have 1120 grams of solution.

$1 lts solution * \frac{1,12 grs solution}{mL solution}*\frac{1000 mL solution}{1 lts solution} = 1120 grs of solution$

With the molecular weight of solute (Sum of: for carbon = 12*12=144; for hydrogen = 1*22=22 and for oxygen = 16*11=176. Final result = 342 grs per mol), we can obtain the mass of solute:

$1,22 mol solute*\frac{342 grs solute}{1 mol solute} = 417,24 grs solute$

Now, the mass of solvent is: mass solvent = mass of solution - mass of solute. So, we have: 1120 - 417,24 = 702,76 grs of solvent = 0,70276 Kgs of solvent

$molality = m = \frac{1,22 mol solute}{0,70276 kgs solvent}= 1,74 molal$

For b) question we have that the mass percent of solute is hte ratio between the mass of solute and the mass of solution. So,

$%(w/w) = \frac{417,24 grs solute}{1120 grs solution} = 37,2%$

For c) question we have that the mole fraction of solute is the ratio between moles of solute and moles of solution. Let's calculate the moles of solution as follows: Moles solution = moles solute + moles solvent. First we have that the moles of solvent are (remember that the molecular weight of water for this calculus is 18 grs per mol):

$702,76 grs solvent*\frac{1 mol solvent}{18 grs solvent} = 39,04 moles solvent$

So, we have the moles of solution: 1,22 moles of solute + 39,04 moles of solvent = 40,26 moles of solution

Finally, we have:

$Mol frac solute = \frac{1,22 mol solute}{40,26 mol solution}= 0,03$

6 0

3 years ago