Respuesta :
Answer:
(a) 2000 mL
(b) 70 mL, 30 mL
(c) No
Step-by-step explanation:
Given that Artesia is a 70% acid solution, and Borromeo is a 20% acid solution.
(a) Let x mL of Artesia is added to 500 mL of Borromeo to produce a 60% acid solution.
So, 70% of x + 20% of 500 = 60% of (x+500)
[tex]\Rightarrow 0.7 \times x + 0.2 \times 500 = 0.6 \times (x+500) \\\\\ \Rightarrow 0.7 \times x + 0.2 \times 500 = 0.6 \times x+0.6 \times500\\\\ \Rightarrow 0.7 \times x -0.6 \times x= 0.6 \times500- 0.2 \times 500 \\\\ \Rightarrow (0.7-0.6) \times x = (0.6-0.2) \times500\\\\\Rightarrow 0.1 \times x = 0.4 \times500\\\\[/tex]
[tex]\Rightarrow x = (0.4 \times500)/0.1=2000[/tex] mL.
(b) Let x mL of Artesia and y mL be combined in order to form a 100 mL solution having 55% acid solution.
So, x+y=100\Rightarrow y= 100-x
And, 70% of x + 20% of y = 55% of (100)\\\\
[tex]\Rightarrow 0.7 \times x + 0.2 \times (100-x) = 0.55 \times 100\\\\\Rightarrow 0.7 \times x + 0.2 \times 100- 0.2 \times x = 0.55 \times 100\\\\\Rightarrow 0.7 \times x -0.2 \times x= 0.55 \times100- 0.2 \times 100 \\\\\Rightarrow (0.7-0.2) \times x = (0.55-0.2) \times100\\\\\Rightarrow 0.5 \times x = 0.35 \times100\\\\[/tex]
[tex]\Rightarrow x = (0.35 \times100)/0.5=70[/tex] mL.
and y= 100-x= 100-70=30 mL.
(c) Let x mL of Artesia and y mL of Borromeo be combined in order to make 80% acid solution.
And, 70% of x + 20% of y = 80% of (x+y)
[tex]\Rightarrow 0.7 \times x + 0.2 \times y= 0.8 \times (x+y)\\\\\Rightarrow 0.7 \times x + 0.2 \times y = 0.8 \times x +0.8 \times y\\\\\Rightarrow (0.7-0.8) \times x = (0.8-0.2) \times y\\\\\Rightarrow -0.1 \times x = 0.6 \times y \\\\\Rightarrow \frac {x}{y} = -6 \\\\[/tex]
As x and y are amount of Artesia and Borromeo, so x>0 and y>0
[tex]\Rightarrow x/y >0[/tex], so the negative value of x/y is not possible.
Ad the negative value of x/y is not so x/y=-6 is not possible, hence, any combination of Artesia and Borromeo that is 80% acid solution is not possible.
