All categories

3 years ago

The correct mathematical expression for finding the molar solubility (s) of sn(oh)2 is:

1 answer:

7 0

Answer is: Ksp = 4s³.
Balanced chemical reaction (dissociation) of strontium hydroxide:
Sr(OH)₂(s) → Sr²⁺(aq) + 2OH⁻(aq).
Ksp(Sr(OH)₂) = [Sr²⁺]·[OH⁻]².
[Sr²⁺] = s.
[OH⁻] = [Sr²⁺] = 2s
Ksp = (2s)² · x = 4s³.
Ksp is the solubility product constant for a solid substance dissolving in an aqueous solution.
[Sr²⁺] is equilibrium concentration of iumcations.
[OH⁻] is equilibrium concentration of hydroxide anions.

You might be interested in

An oxide of phosphorus contains 56.4% phosphorus and 43.6% oxygen. It's relative molecular mass is 220. Find both the empirical

sladkih [1.3K]

Moles of P = 56,4g/30,974g/mole = 1,82 moles P
moles of O = 43,6/15,999 = 2,73 moles of O

converting to the simplest ratio:
For P : 1,82/1,82 = 1
For O : 2,73/1,82 = 1,5

1 P and 2 oxygens.
PO2 -> the empirical formula

hope this help

7 0

3 years ago

Read 2 more answers

Tartaric acid, C4H6O6, has the first ionization constant with the value: Ka1 = 9.20 × 10-4. Calculate the value of pKb for the c

nirvana33 [79]

Answer:

pKb = 10.96

Explanation:

Tartaric acid is a dyprotic acid. It reacts to water like this:

H₂Tart + H₂O ⇄ H₃O⁺ + HTart⁻ Ka1

HTart⁻ + H₂O ⇄ H₃O⁺ + Tart⁻² Ka2

When we anaylse the base, we have

Tart⁻² + H₂O ⇄ OH⁻ + HTart⁻ Kb1

HTart⁻ + H₂O ⇄ OH⁻ + H₂Tart Kb2

Remember that Ka1 . Kb2 = Kw, plus pKa1 + pKb2 = 14

Kb2 = Kw / Ka1 → 1×10⁻¹⁴ / 9.20×10⁻⁴ = 1.08×10⁻¹¹

so pKb = - log Kb2 → - log 1.08×10⁻¹¹ = 10.96

7 0

3 years ago

Read 2 more answers

Give the values of n, l and ml for: (a) each orbital in the 2p subshell, (b) each orbital in the 5d subshell.?

SCORPION-xisa [38]

Hey there!:

* For 2p subshell :

n = 2, l =1, ml = -1, 0, +1

* for 5d subshell,

n = 5, l = 2, ml = -2, -1, 0, +1, +2

Hope that helps!

7 0

3 years ago

The bond between sodium and chlorine in the compound sodium chloride (NaCl) is a/an:

Deffense [45]

The classic case of ionic bonding, the sodium chloride molecule forms by the ionization of sodium and chlorine atoms and the attraction of the resulting ions. An atom of sodium has one 3s electron outside a closed shell, and it takes only 5.14 electron volts of energy to remove that electron.

8 0

3 years ago

A particular laser consumes 130.0 Watts of electrical power and produces a stream of 2.67×1019 1017 nm photons per second.

solniwko [45]

The missing question is:

What is the percent efficiency of the laser in converting electrical power to light?

The percent efficiency of the laser that consumes 130.0 Watt of electrical power and produces a stream of 2.67 × 10¹⁹ 1017 nm photons per second, is 1.34%.

A particular laser consumes 130.0 Watt (P) of electrical power. The energy input (Ei) in 1 second (t) is:

$Ei = P \times t = 130.0 J/s \times 1 s = 130.0 J$

The laser produced photons with a wavelength (λ) of 1017 nm. We can calculate the energy (E) of each photon using the Planck-Einstein's relation.

$E = \frac{h \times c }{\lambda }$

where,

h: Planck's constant

c: speed of light

$E = \frac{h \times c }{\lambda } = \frac{6.63 \times 10^{-34}J.s \times 3.00 \times 10^{8} m/s }{1017 \times 10^{-9} m }= 6.52 \times 10^{-20} J$

The energy of 1 photon is 6.52 × 10⁻²⁰ J. The energy of 2.67 × 10¹⁹ photons (Energy output = Eo) is:

$\frac{6.52 \times 10^{-20} J}{photon} \times 2.67 \times 10^{19} photon = 1.74 J$

The percent efficiency of the laser is the ratio of the energy output to the energy input, times 100.

$Ef = \frac{Eo}{Ei} \times 100\% = \frac{1.74J}{130.0J} \times 100\% = 1.34\%$

The percent efficiency of the laser that consumes 130.0 Watt of electrical power and produces a stream of 2.67 × 10¹⁹ 1017 nm photons per second, is 1.34%.

You can learn more about lasers here: brainly.com/question/4869798

8 0

2 years ago