Barium-131's radiation level won't reach 1/4 of its initial level for 24 hours.
ln[A] t = -kt + ln[A] 0 is the integrated rate rule for the first-order reaction A's products.
A straight line is produced when the natural log of [A] is plotted as a function of time since this equation has the form y = mx + b.
How is the length of a half-life determined?
The amount of time needed for the reactant concentration to drop to half its initial value is known as the half-life of a reaction. A first-order reaction's half-life is a constant that is correlated with its rate constant:
t 1/2 = 0.693/k.
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In 1 mol of CH3OH, you have 4 H-atoms (because 3 H-atoms
are attached to the C-atom, and one H-atom in the OH group). That means
in 0.500 mol of CH3OH, you have 2 H-atoms since it is halved. And then we have Avogadro's constant: 6.02 * 1023.
The question asks for how many hydrogen atoms there are in 0.500 mol CH3OH. Using the numbers that we have (Avogadro's constant and no. of H-atoms), the answer of the question will be something like:
<span>H-atoms in CH3OH = 2 * 6.02 * </span>1023<span> = ~1.2 * 10</span>24
<span>V equals one-third times pi times r squared times h</span>
