I'm preatty sure it is in the periodic table xxx
6N
Explanation:
you times 3 and 2 to get six.
<u>Answer: </u>
A sample initially contained 150 mg of radon-222. After 11.4 days only 18.75mg of the radon-222 in the sample remained where 3 half-lives have passed
<u>Explanation:</u>
Given, the initial value of the sample, = 150mg
Final value of the sample or the quantity left, A = 18.75mg
Time = 11.4 days
The amount left after first half life will be ½.
The number of half-life is calculated by the formula
where N is the no. of half life
Substituting the values,
On equating, we get, N = 3
Therefore, 3 half-lives have passed.
Aldol condensation involves two aldehydes or two ketones or an aldehyde and a ketone. The product of the reaction is shown in the image attached.
<h3>What is Aldol condensation?</h3>
Aldol condensation is a reaction that involves two aldehydes or two ketones or an aldehyde and a ketone. This reaction occurs at the carbonyl group as one of the carbonyl compounds is in the enolate form.
The product of the reaction is shown in the image attached. This gives the mechanism for the Aldol condensation of the compound 3‑methylbutanal.
Learn more about aldol condensation: brainly.com/question/9415260
The silver ion concentration in saturated solution of silver (i) phosphate is calculated as follows.
write the equation for dissociation of silver (i) phosphate
that is Ag3PO4 (s) = 3Ag^+(aq) + PO4 ^3-(aq)
let the concentration of the ion be represented by x
ksp is therefore= (3x^3 )(x) = 1.8 x10 ^-18
27 x^3 (x) = 1.8 x10^-18
27x^4 = 1.8 x10^-18 divide both side 27
X^4 = 6.67 x10 ^-20
find the fourth root x = 1.6 x10 ^-5m
the concentration of silver ion is therefore = 3 x (1.6 x10^-5) = 4.8 x10^-5m