Hihi!
Properties of an electromagnetic wave can be amplitude<span>, wavelength, and frequency!</span>
I hope I helped!
-LoliArual
<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 molarity of the HCl solution should be 4.04 M
Explanation:
<u>Step 1:</u> Data given
volume of HCl solution = 10.00 mL = 0.01 L
volume of a 1.6 M NaOH solution = 25.24 mL = 0.02524 L
<u>Step 2:</u> The balanced equation
HCl + NaOH → NaCL + H2O
Step 3: Calculate molarity of HCl
n1*C1*V1 = n2*C2*V2
Since the mole ratio for HCl and NaOH is 1:1 we can just write:
C1*V1 =C2*V2
⇒ with C1 : the molarity of HCl = TO BE DETERMINED
⇒ with V1 = the volume og HCl = 10 mL = 0.01 L
⇒ with C2 = The molarity of NaOH = 1.6 M
⇒ with V2 = volume of NaOH = 25.24 mL = 0.02524 L
C1 * 0.01 = 1.6 * 0.02524
C1 = (1.6*0.02524)/0.01
C1 = 4.04M
The molarity of the HCl solution should be 4.04 M
Calcium Chloride is the salt that is effective for melting ice on roads
