The Answer is a b¹⁰
Simplify the following:
(a^5 b^6 b^4)/a^4
Combine powers. (a^5 b^6 b^4)/a^4 = a^(5 - 4) b^(6 + 4):
a^(5 - 4) b^(6 + 4)
5 - 4 = 1:
a b^(6 + 4)
6 + 4 = 10:
Answer: a b^10
The question is incomplete, here is the complete question:
The half-life of a certain radioactive substance is 46 days. There are 12.6 g present initially.
When will there be less than 1 g remaining?
<u>Answer:</u> The time required for a radioactive substance to remain less than 1 gram is 168.27 days.
<u>Step-by-step explanation:</u>
All radioactive decay processes follow first order reaction.
To calculate the rate constant by given half life of the reaction, we use the equation:
where,
= half life period of the reaction = 46 days
k = rate constant = ?
Putting values in above equation, we get:
The formula used to calculate the time period for a first order reaction follows:
where,
k = rate constant =
t = time period = ? days
a = initial concentration of the reactant = 12.6 g
a - x = concentration of reactant left after time 't' = 1 g
Putting values in above equation, we get:
Hence, the time required for a radioactive substance to remain less than 1 gram is 168.27 days.
We can get the measurement of the arc by determining first the radius of the circle. The circumference of a circle is equal to 2 * pi * radius. Hence from a 12 unit2 circumferenced circle, the radius is 1.91 units.Length of the arc is equal to radius times the angle in radians. 60 deg is equal to pi over 3. The length of the arc is equal to 2 units.
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step by step explanation: