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serious [3.7K]
4 years ago
13

An aqueous solution at 25°c has a poh of 7.87 . calculate the ph. be sure your answer has the correct number of significant digi

ts.
Chemistry
2 answers:
lorasvet [3.4K]4 years ago
5 0
Answer is: a<span>n aqueous solution at 25°C has a pH of 6,13.
</span>pOH(solution) = 7,87
pH + pOH = 14.
pOH(solution) = 14 - 7,87
pOH(solution) = 6,13.
pH<span> is the negative of the logarithm of the H</span>3O⁺<span> ion concentration.
</span>pOH<span> is the negative of the logarithm of the OH</span>⁻ <span>ion concentration.</span>
Kamila [148]4 years ago
4 0

Answer:

pH = 6.13 (3 significant figures)

Explanation:

<u>Given:</u>

pOH of the solution = 7.87

<u>To determine:</u>

pH of the solution

<u>Explanation:</u>

pH is a parameter which reflects the H+ ion concentration (acidity) of a given solution where pOH is indicative of the OH- ion (basicity) concentration. The two are related as follows:

pH + pOH = 14.0\\\\Therefore, pH = 14.0-pOH

For the given solution:

pH = 14.0 - 7.87 = 6.13

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Sav [38]

Explanation:

The substance that is in excess that doesn't get used up as a reactant is called limiting reactant.

7 0
3 years ago
One degree Celsius indicates the same temperature change as
Harlamova29_29 [7]
B 5/9 degree Fahrenheit
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4 years ago
The normal boiling point of methanol is 64.7∘c and the molar enthalpy of vaporization of 71.8kj/mol. the value of δs when 1.11 m
maxonik [38]
The answer is 236.5 J/K

According to Δ G formula:

ΔG = ΔH - TΔS

when ΔG is the change in free energy (KJ)

and ΔH is the change in enthalpy (KJ)= ΔHvap * moles

                                                              = 71.8 KJ/mol * 1.11 mol
                                                             
                                                              =   79.7 KJ

and T is the absolute temperature (K)= 64 °C + 273°C = 337 K

Δ S is the change in entropy  KJ/K

by substitution:

when at equilibrium ΔG = 0 

∴ΔS = ΔH / T

       =79.7 KJ/ 337 K

     = 0.2365 KJ/K

     = 236.5 J/K


6 0
3 years ago
A solution containing CaCl2 is mixed with a solution of Li2C2O4 to form a solution that is 3.5 × 10-4 M in calcium ion and 2.33
Burka [1]

Answer:

B. A precipitate will form since Q > Ksp for calcium oxalate

Explanation:

Ksp of CaC₂O₄ is:

CaC₂O₄(s) ⇄ Ca²⁺ + C₂O₄²⁻

Where Ksp is defined as the product of concentrations of Ca²⁺ and C₂O₄²⁻ in equilibrium:

Ksp = [Ca²⁺][C₂O₄²⁻] = 2.27x10⁻⁹

In the solution, the concentration of calcium ion is 3.5x10⁻⁴M and concentration of oxalate ion is 2.33x10⁻⁴M.

Replacing in Ksp formula:

[3.5x10⁻⁴M][2.33x10⁻⁴M] = 8.155x10⁻⁸. This value is reaction quotient, Q.

If Q is higher than Ksp, the ions will produce the precipitate CaC₂O₄ until [Ca²⁺][C₂O₄²⁻] = Ksp.

Thus, right answer is:

<em>B. A precipitate will form since Q > Ksp for calcium oxalate</em>

<em></em>

4 0
3 years ago
How many are molecules ( or formula) in each sample?
andre [41]

Answer:

  • 4.010 \times 10^{25} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}
  • 16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}

<u>Explanation</u>:

<u>Number of molecules for 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}</u>

\text { Firstly molar mass is calculated of } \mathrm{NaHCO}_{3}:

Atomic mass of Na + H + C + 3(O)  = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol

\text { Number of molecules of } \mathrm{NaHCO}_{3} \text { in } 55.93 \text { kg are as follows: }

55.93 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} \mathrm{NaHCO}_{3}}{84.00 \mathrm{gm} \mathrm{NaHCO}_{3}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number }\right)

=4.010 \times 10^{26} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}

<u>Number of molecules for for \left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}</u>

\text { Firstly molar mass is calculated of } \mathrm{Na}_{3} \mathrm{PO}_{4}

= Atomic mass of 3(Na) + P + 4(O)

= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

459 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} N a_{3} P O_{4}}{163.94 \mathrm{gm} N a_{3} P O_{4}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number) } / 1 \mathrm{mol}\right.

=16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}

8 0
3 years ago
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