<span>85% ethanol | 25% ethanol | 50% ethanol
x | y | 20 gal
use x and y because you don;t know how much she needs.
0.85x | 0.25y | 20(0.5)
85% is 85/100 or 0.85, and you need that much of x, same goes for the 25% and 50% mixtures so now you can make up 2 equations
1) x + y = 20 2) 0.85x + 0.25y= 10 (you get 10 when you multiply 20 by 0.5) now you can solve for x or y using substitution.
first rewrite 1) in terms of x or y: x+ y= 20 ----> y= 20 - x now you can substitute 20- x for y in the second equation.. 0.85x + 0.25y= 10 0.85x + 0.25(20-x)= 10 distribute here..(0.25 * 20 and 0.25 * (-x) ) 0.85x + 5 - 0.25x = 10 combine like terms 0.6x +5 = 10 move the 5 over to the other side 0.6x= 10 -5 0.6x = 5 divide both sides by 0.6 x= 25/3 or 8.3 now you know the amount of x so you can substitue this back into the first equation to find y. 0.85x + 0.25y= 10 0.85(25/3) +0.25y= 10 85/12 + 0.25y= 10 0.25y = 10- 85/12 0.25y= 35/12 y= 35/3 or 11.6 you can check by putting these values into the euations: 1) x+ y= 20 25/3 + 35/3 =20 20= 20 good so far 2) 0.85x + 0.25y= 10 0.85(25/3) + 0.25(35/3)=10 10 = 10
so our values for x and y work
x= 25/3 and y= 35/3</span>
Answer: the area of the telescope lens is \textit{0,08507 ft}
Explanation: Happy I could help!
Answer:
chromium:oxygen=1:2
Explanation:
total mass=2.16g
mass of chromium=1.34g
mass of oxygen=2.16g-1.34g=0.82g
chromium oxygen
mass:1.34g 0.82
RAM:52 16
divide the mass with the RAM
0.02577 0.05125
divide by smallest number to find ratio
1 2
ratio of chromium to oxygen=1:2
Answer:
0.124 M
Explanation:
The reaction obeys second-order kinetics:
![r = k[BrO^-]^2](https://tex.z-dn.net/?f=r%20%3D%20k%5BBrO%5E-%5D%5E2)
According to the integrated second-order rate law, we may rewrite the rate law in terms of:
![\dfrac{1}{[BrO^-]_t} = kt + \dfrac{1}{[BrO^-]_o}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B%5BBrO%5E-%5D_t%7D%20%3D%20kt%20%2B%20%5Cdfrac%7B1%7D%7B%5BBrO%5E-%5D_o%7D)
Here:
is a rate constant,
is the molarity of the reactant at time t,
is the initial molarity of the reactant.
Converting the time into seconds (since the rate constant has seconds in its units), we obtain:

Rearranging the integrated equation for the amount at time t:
![[BrO^-]_t = \dfrac{1}{kt + \dfrac{1}{[BrO^-]_o}}](https://tex.z-dn.net/?f=%5BBrO%5E-%5D_t%20%3D%20%5Cdfrac%7B1%7D%7Bkt%20%2B%20%5Cdfrac%7B1%7D%7B%5BBrO%5E-%5D_o%7D%7D)
We may now substitute the data:
![[BrO^-]_t = \dfrac{1}{0.056 M^{-1}s^{-1}\cdot 60.0 s + \dfrac{1}{0.212 M}} = 0.124 M](https://tex.z-dn.net/?f=%5BBrO%5E-%5D_t%20%3D%20%5Cdfrac%7B1%7D%7B0.056%20M%5E%7B-1%7Ds%5E%7B-1%7D%5Ccdot%2060.0%20s%20%2B%20%5Cdfrac%7B1%7D%7B0.212%20M%7D%7D%20%3D%200.124%20M)
1.53 moles of Fe is your solution hope it helps!