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max2758
You are working in a chemistry lab. You have 1000 milliliters of pure acid. A dilution of acid is created by adding pure acid to water. A 40% dilution contains 40% acid and 60% water. You have been asked to make a 40% dilution and a 60% dilution of pure acid. a.Write an equation in standard form that models the possible quantities of each dilution you can prepare using all 1000 milliliters of pure acid. b.You prepare 700 milliliters of the 40% dilution how much of the 60% dilution can you prepare c. How much water do you need to prepare 700 milliliters of the 40% dilution?
\[A)\ Dilution=\frac{V_a}{V_a+V_w}\]in percent, where V_a is equal to 1000(ml) \[B)\ \ \ 0.4=\frac{V_a}{700}\] \[V_a = 280\] I have 1000-280 = 720 ml of acid left \[0.6 = \frac{720}{720+V_w}\] \[V_w = 480\] I can prepare 720+480 = 1200 milliliters of the60% dilution \[C)\ \ \ 0.4=\frac{V_a}{700} \] \[V_a = 280\] \[V_w = 700 - 280 = 420\] I need 420ml of water to prepare the 700ml 40% dilution
so thats the equation for part a
It hass tobe in standard form
\[Dilution(V_a + V_w)=V_a\] \[Dilution*V_a + Dilution *V_w = V_a\] Where Dilution is a positive rational.
so does it change for all the other parts b and c
Hmm... wait my answer isn't in standard form
Thats not in standard form Standard form is like this 2x+2y=200020202020
thats just an example though
I can write it as \[V_w+(1-\frac{1}{a})*V_a=0\] which kinda matches the standard form\[Ax + By = C\]where A is 1, B is (1-1/a), and C is 0
Just put it in standard form like right now Ax+By=c
It is in standard form
Just write it again in standard form.
\[V_w + (1-\frac{1}{a})*V_a=0\]
You may substitute x for V_w, and y for V_a\[x + (1-\frac{1}{a})y=)\]
oops:\[x+(1-\frac{1}{a})y = 0\]
Thanks ok i have one more question its at top can you help me with it its not as hard as this one