<span>What
is the ph of an acetic acid solution if 10 drops are titrated with 70
drops of a 0.65 m koh solution? (ka for acetic acid = 1.8 x 10-5)?
[KOH] = 0.65 M
[OH] = 0.65 M
</span>Dilute your mom
<span>[OH]Dil= 0.65 M * 70/80 = 0.56875 M
pH = 5.4
</span>
The answer is 3 It is a body of knowledge gained using inquiry and experimentation. Hope this helped!
Answer:
32.4 mol
Explanation:
Given data:
Number of moles of C atom present = ?
Number of moles of glucose = 5.4 mol
Solution:
Glucose formula = C₆H₁₂O₆
There are 6 moles of C atoms are present in one mole of glucose.
In 5.4 moles of glucose:
5.4 mol × 6 = 32.4 mol
Answer:
6. d, 7. a
Explanation:
6. Molarity is a number of moles solute in 1 L solution.
7. 1 L solution - 2.5 mol K2CO3
20 L - x mol K2CO3
x =20*2.5/1 = 50 mol K2CO3
Molar mass(KCO3) = M(K) + M(C) + 3M(O)= 39 +12 +3*16= 99 g/mol
99 g/mol *50 mol = 4950 g KCO3 Closest answer is A.
Actually KCO3 does not exist, in reality it should be K2CO3.
