Answer:
0.00370 g
Explanation:
From the given information:
To determine the amount of acid remaining using the formula:
where;
v_1 = volume of organic solvent = 20-mL
n = numbers of extractions = 4
v_2 = actual volume of water = 100-mL
k_d = distribution coefficient = 10
∴
Thus, the final amount of acid left in the water = 0.012345 * 0.30
= 0.00370 g
%C= 12/12 + 2·16=0,273=27,3%.
Answer:
a. 2,9x10⁻⁴ M HCl
Explanation:
A solution is considered acidic when its concentration of H⁺ is higher than 1x10⁻⁷. The higher concentration of H⁺ will be the most acidic solution.
a. 2,9x10⁻⁴ M HCl. In water, this solution dissolves as H⁺ and Cl⁻. That means concentration of H⁺ is 2,9x10⁻⁴ M.
b. 4,5x10⁻⁵M HNO₃. In the same way, concentration of H⁺ is 4,5x10⁻⁵M.
c. 1,0x10⁻⁷M NaCl. As this solution doesn't produce H⁺, the solution is not acidic
d. 1,5x10⁻²M KOH. This solution produce OH⁻. That means the solution is basic nor acidic.
Thus, the solution considered the most acidic is a. 2,9x10⁻⁴ M HCl, because has the higher concentration of H⁺.
I hope it helps!
Answer:
a) a0 was 46.2 grams
b) It will take 259 years
c) The fossil is 1845 years old
Explanation:
<em>An unknown radioactive substance has a half-life of 3.20hours . If 46.2g of the substance is currently present, what mass A0 was present 8.00 hours ago?</em>
A = A0 * (1/2)^(t/h)
⇒ with A = the final amount = 46.2 grams
⇒ A0 = the original amount
⇒ t = time = 8 hours
⇒ h = half-life time = 3.2 hours
46.2 = Ao*(1/2)^(8/3.2)
Ao = 261.35 grams
<em>Americium-241 is used in some smoke detectors. It is an alpha emitter with a half-life of 432 years. How long will it take in years for 34.0% of an Am-241 sample to decay?</em>
t = (ln(0.66))-0.693) * 432 = 259 years
It will take 259 years
<em>A fossil was analyzed and determined to have a carbon-14 level that is 80% that of living organisms. The half-life of C-14 is 5730 years. How old is the fossil?</em>
<em />
t = (ln(0.80))-0.693) * 5730 = 1845
The fossil is 1845 years old
That question shoulda be true
