If x+4 is a factor of x^2-ax-12a and a is a constant, what is the value of a?
1 answer:
You might be interested in
Based on the Ksp calculations, the molar solubility of AgCl in 0.500 M of NH₃ is equal to 2.77 × 10⁻² M.
<h3>How to determine the molar solubility.</h3>First of al, we would write the properly balanced chemical equation for this chemical reaction:
AgCl(s) ⇆ Ag⁺(aq)+ Cl⁻(aq) Ksp = 1.8 × 10⁻¹⁰
AgCl(s) + 2NH₃(aq) ⇆ Ag(NH₃)₂⁺(aq) Kf = 1. 7 × 10⁷
<u>Given the following data:</u>
- Ksp of AgCl = 1.8 × 10⁻¹⁰
- Concentration of NH₃ = 0.500
- Kf of ag(nh₃)₂⁺ = 1.7 × 10⁷
K = Ksp × Kf
K = 1. 80 × 10⁻¹⁰ × 1.7 × 10⁷
K = 3.06 × 10⁻³
Mathematically, the Ksp for the above chemical reaction is given by:
x = 2.77 × 10⁻² M.
Read more on molar solubility here: brainly.com/question/3006391
<h3>Answer:</h3>
60 days
<h3>Explanation:</h3>Half-life is the amount of time it takes half of a substance to decay away.
Guess and Check
One method for solving half-life problems like this is to guess and check. To do this, we can continue to divide 475mg by 2 until we get to 30mg.
- 475 ÷ 2 = 237.5
- 237.5 ÷ 2 = 118.75
- 118.75 ÷ 2 = 59.375
- 59.375 ÷ 2 = 29.6875
As seen here, it takes approximately 4 half-lives for this sample of Ra-225 to decay to 30mg. Now, we can multiply 4 by the length of the half-life, 15 days.
- 4 * 15 = 60 days.
Fractions
Another way to solve this is to use fractions.
- 30 is about 1/16 of 475
- 16 is equivalent to
.
This means that it takes 4 half-lives for 475mg to decay to 30mg. Using the same method above, we can tell that 4 half-lives are 60 days.
Important Note
In this question, I rounded occasionally. So, not all of the values are exact, but they are all very close.