Concentration of unknown acid is 0.061 M
Given:
Concentration of NaOH = 0.125 M
Volume of NaOH = 24.68 mL
Volume of acid solution = 50.00 mL
To Find:
Concentration of the unknown acid
Solution: Concentration is the abundance of a constituent divided by the total volume of a mixture. The concentration of the solution tells you how much solute has been dissolved in the solvent
Here we will use the formula for concentration:
M1V1 = M2V2
0.125 x 24.68 = 50 x M2
M2 = 0.125 x 24.68 / 50
M2 = 0.061 M
Hence, the concentration of unknown acid is 0.061 M
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Answer:
the answer is B.
No, as ionic compounds are only conductive in an aqueous (water) solution
Explanation: i just know
Answer:
There were originally 8 atoms of Potassium-40.
Explanation:
The half-life of a radioactive material is the time taken for half the original material to decay or the time required for a quantity of the radioactive substance to reduce to half of its initial value.
If the original material formed without any Argon-40, it means that the atoms originally present were Potassium-40 atoms.
Presently, there are 7 Argon-40 atoms for every 1 of Potassium-40, we can deduce the number of half-lifes the Potassium-40 has undergone as follows :
After one half-life, (1/2) there will be one Potassium-40 atom for every Argon-40 atom.
After a second half life, 1/2 × 1/2 = 1/4: there will be one Potassium-40 atom for every three atoms of Argon-40.
After a third half-life, 1/4 × 1/2 = 1/8: there will be one Potassium-40 atom for every 7 atoms of Argon-40.
Since there are 1/8 atoms of Potassium-40 presently, there were originally 8 atoms of Potassium-40.
Answer:
1- 0.04 M/s.
2- 0.16 M/s.
Explanation:
- For the reaction: 4PH₃ → P₄ + 6H₂.
<em>The rate of the reaction = - d[PH₃]/4dt = d[P₄]/dt = d[H₂]/6dt.</em>
where, - d[PH₃]/dt is the rate of PH₃ changing "rate of disappearance of PH₃".
d[P₄]/dt is rate of P₄ changing "rate of appearance of P₄".
d[H₂]/dt is the rate of H₂ changing "rate of formation of H₂" (d[H₂]/dt = 0.24 M/s).
<u><em>(a) At what rate is P₄ changing?</em></u>
∵ The rate of the reaction = d[P₄]/dt = d[H₂]/6dt.
∴ <em>rate of P₄ changing = </em>d[P₄]/dt = d[H₂]/6dt = (0.240 M/s)/(6.0) = 0.04 M/s.
<u><em>(b) At what rate is PH</em></u>₃<u><em> changing?</em></u>
∵ The rate of the reaction = - d[PH₃]/4dt = d[H₂]/6dt.
∴ <em>rate of PH</em>₃<em> changing = </em>- d[PH₃]/dt = 4(d[H₂]/6dt) = (4)(0.240 M/s)/(6.0) = 0.16 M/s.
