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2 years ago

14

At 25 °C, an aqueous solution has an equilibrium concentration of 0.00423 0.00423 M for a generic cation, A + A+ (aq), and 0.004

23 0.00423 M for a generic anion, B − B− (aq). What is the equilibrium constant, K sp Ksp , of the generic salt AB (s)

1 answer:

kow [346]2 years ago

4 0

Answer : The value of $K_{sp}$ of the generic salt AB is, $1.79\times 10^{-5}$

Explanation : Given,

Concentration of generic cation, $A^+$ = 0.00423 M

Concentration of generic anion, $B^-$ = 0.00423 M

The equilibrium chemical reaction will be:

$AB(s)\righleftharpoons A^+(aq)+B^-(aq)$

The solubility constant expression for this reaction is:

$K_{sp}=[A^+][B^-]$

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

$K_{sp}=(0.00423M)\times (0.00423M)$

$K_{sp}=1.79\times 10^{-5}$

Thus, the value of $K_{sp}$ of the generic salt AB is, $1.79\times 10^{-5}$

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Answer:

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Explanation:

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2 years ago

If a 0.4856 gram sample of khp is dissolved in sufficient water to prepare 250 ml of solution, and 25 ml of the solution require

dangina [55]

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$n=\frac{m}{M}$

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$n=\frac{0.4856 g}{204.22 g/mol}=0.00237 mol$

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Detailed calculations:

Molarity is defined as number of moles in 1 L of solution, for 250 mL of solution, molarity will be:

$M=\frac{0.00237 mol}{250 \times 10^{-3}L}=0.009511 M$

For 25 mL, apply dilution law as follows:

$M_{1}V_{1}=M_{2}V_{2}$

Putting the values,

$0.009511\times 250=M_{2}\times 25 mL$

On rearranging,

$M_{2}=\frac{0.009511\times 250}{25}=0.09511 M$

Convert molarity into number of moles,

$n=M\times V=0.09511 mol/L\times 25\times 10^{-3}L=0.00237 mol$

At equivalent point, number of moles of KHP will be equal to NaOH, thus, number of moles of NaOH will be 0.00237 mol.

Calculation for molarity:

Volume of NaOH is 18.75 mL, thus, molarity can be calculated as follows:

$M=\frac{n}{V}$

Putting the values,

$M=\frac{0.00237 mol}{18.75\times 10^{-3}L}=0.1264 M$

Therefore, molarity of NaOH is 0.1264 M

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2 years ago

When an electron in a 2p orbital of a particular atom makes a transition to the 2s orbital, a photon of approximate wavelength 6

Mariulka [41]

Answer:

The energy difference between these 2p and 2s orbitals is $3.07\times 10^{-19} J$

Explanation:

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Energy of the photon will corresponds to the energy difference between 2p and 2s orbital = E

Energy of the photon is given by Planck's equation:

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

h = Planck's constant = $6.626\tiomes 10^{-34} Js$

c = Speed of the light = $3\times 10^8 m/s$

$E=\frac{6.626\tiomes 10^{-34} Js\times 3\times 10^8 m/s}{646.3\times 10^{-9} m}$

$E=3.07\times 10^{-19} J$

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2 years ago

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Taya2010 [7]

Answer:

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We have, $T_c=20^{\circ} C$

So,

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So, the room temperature is 293.15 kelvin.

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seropon [69]

Answer:

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Step-by-step explanation:

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2 years ago