help with math please
The question is..
Suppose that x ml of pure acid is added to 130ml of a 40% acid solution. Express the concentration C(x) of the new mixture as a function of x
I would have written down 0.4+1/130+x... but the correct answer is 0.4(130) + x(1)/130+x
can someone explain why I am wrong?
November 4, 2012
8 Comments • Newest first
[quote=LowWillpower]C(x) is going to be 0.4 + x/(130 + x), 0.4*(130) is not a concentration, it's an amount (40% of 130 mL), I don't see why concentration * volume = concentration.[/quote]
Woops, my bad, you're right, the concentration is just going to be 0.4.
[quote=CaptCandy]No, you wrote the solution wrong, it is: C(x) = 0.4(130) + x(1)/(130+x)
Because the current concentration is the 0.4 part and for every x ml of pure acid, hence x's coefficient is 1, you increase acid concentration by that amount, but you also have to divide by an increase in volume, which is the quantity (130 + x), the new volume.
In a simpler format:
- C(x) is the concentration of acid
- 0.4(130) is the original concentration and volume of the mixture. This is the initial concentration
- x(1) is the increase in ml of acid per x ml added to the mixture
- (130+x) is the denominator representing the new volume after x ml of acid is added to the mixture. When x(1) is divided by this quantity, it results in additional concentration of acid per x ml of acid added to the mixture.
Thus, the equation above represents original acid concentration + dc/dx, or change in concentration per change in x.
Now in Calc II, it gets more complicated where you have to integrate a function that expresses the rate at which the x ml is flowing into the pool.[/quote]
C(x) is going to be 0.4 + x/(130 + x), 0.4*(130) is not a concentration, it's an amount (40% of 130 mL), I don't see why concentration * volume = concentration.
[quote=CaptCandy]No, ml is just the unit of x, it is not multiplied. For instance, 10kg is multiplied by 2m/s^2 to get 20 Newtons is written as 10kg * 2m/s^2 = 20N[/quote]
ohh kk thanks makes moar sense now.
[quote=akibari]but why are you multiplying the ml with the concentration? (the .4 x 130) ><? I am legit r-tarded atm D: .[/quote]
No, ml is just the unit of x, it is not multiplied. For instance, 10kg is multiplied by 2m/s^2 to get 20 Newtons is written as 10kg * 2m/s^2 = 20N
And no, it's ok to make mistakes on the first round, as it means that you have room for improvement.
I edited my post to make it more legible so I hope you can understand it and improve your math skills
I remember back in 5th grade in '05, I didn't know how to compute the area of a triangle and my little sister of two years junior had to help me, so don't feel bad.
[quote=CaptCandy]No, you wrote the solution wrong, it is: 0.4(130) + x(1)/(130+x)
Because the current concentration is the 0.4 part and for every x ml of pure acid, hence x's coefficient is 1, you increase acid concentration by that amount, but you also have to divide by an increase in volume, which is the quantity (130 + x), the new volume.
Now in Calc II, it gets more complicated where you have to integrate a function that expresses the rate at which the x ml is flowing into the pool.[/quote]
but why are you multiplying the ml with the concentration? (the .4 x 130) ><? I am legit r-tarded atm D: .
you first need to correct your algebraic terms. because you did not clearly parenthesize your own equation, we must assume that your answer was "(0.4) + (1/130) + (x)" despite the fact that we know you meant "(0.4 + 1)/(130 + x)"
No, you wrote the solution wrong, it is: C(x) = 0.4(130) + x(1)/(130+x)
Because the current concentration is the 0.4 part and for every x ml of pure acid, hence x's coefficient is 1, you increase acid concentration by that amount, but you also have to divide by an increase in volume, which is the quantity (130 + x), the new volume.
In a simpler format:
- C(x) is the concentration of acid
- 0.4(130) is the original concentration and volume of the mixture. This is the initial concentration
- x(1) is the increase in ml of acid per x ml added to the mixture
- (130+x) is the denominator representing the new volume after x ml of acid is added to the mixture. When x(1) is divided by this quantity, it results in additional concentration of acid per x ml of acid added to the mixture.
Thus, the equation above represents original acid concentration + dc/dx, or change in concentration per change in x.
Now in Calc II, it gets more complicated where you have to integrate a function that expresses the rate at which the x ml is flowing into the pool.
ooh i learned this.
make a chart first with the formula
for ex: R*D=RT
underneath fill in the variables
then create a formula using the chart and solve
i would do this but its dark cuz my mom turned off the lights