For maleic acid, hoocch=chcooh, ka1 = 1.42 10–2 and ka2 = 8.57 10–7 . what is the concentration of maleate ion (–oocch=chcoo
– ) in a 0.150 m aqueous solution of maleic acid?
2 answers:
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Answer:
1.82 g is the maximum mass of CuS.
Explanation:
Considering:
Or,
Given :
<u>For :
</u>
Molarity = 0.500 M
Volume = 38.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 38.0×10⁻³ L
Thus, moles of :
<u>Moles of = 0.019 moles
</u>
<u>For :
</u>
Molarity = 0.600 M
Volume = 42.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 42.0×10⁻³ L
Thus, moles of :
<u>Moles of = 0.0252 moles
</u>
According to the given reaction:
1 mole of reacts with 1 mole of
So,
0.019 mole of reacts with 0.019 mole of
Moles of = 0.019 mole
Available moles of = 0.0252 mole
<u>Limiting reagent is the one which is present in small amount. Thus, is limiting reagent.</u>
The formation of the product is governed by the limiting reagent. So,
1 mole of gives 1 mole of
0.019 mole of gives 0.019 mole of
Moles of formed = 0.019 moles
Molar mass of = 95.611 g/mol
Mass of = Moles × Molar mass = 0.019 × 95.611 g = 1.82 g
<u>1.82 g is the maximum mass of CuS.</u>
Answer:
Explanation:
The half-life of K-40 (1.3 billion years) is the time it takes for half of it to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table as follows:
No. of Fraction
<u>half-lives</u> <u> t/yr </u> <u>Remaining</u>
0 0 1
1 1.3 billion ½
2 2.6 ¼
3 3.9 ⅛
We see that after 2 half-lives, ¼ of the original mass remains.
Conversely, if two half-lives have passed, the original mass must have been four times the mass we have now.
Original mass = 4 × 2.10 g =