To obtain the desired solution:
450 mL of 85% alcohol solution is needed to obtain the desired solution.
Calculation:
Let x be the amount of the 85% alcohol required
The volume of the resulting 70% alcohol solution will then be = x + 135 ml
135 mL of the 20% alcohol solution contains the amount of "pure" alcohol is = 0.20×135 mL.
The 85% alcohol solution contains x mL of "pure" alcohol = 0.85× x mL.
The total amount of the "pure" alcohol is the sum
= 0.20×135 + 0.85× x mL.
It should be equal to the amount of the "pure" alcohol in the mixture, which is = 0.70× (x+135) ml.
So, your "pure alcohol" equation is,
= 0.85× x + 0.20×135 = 0.70× (x+135)
Simplify and solve it for x:
0.85x + 0.20×135 = 0.70x + 0.70×135,
0.85x - 0.70x = 0.70×135 - 0.20×135,
0.15x = 67.5
x = 67.5/0.15
= 450mL.
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Find one mole
8 C = 8 * 12 = 96
9 H = 1 * 9 = 9
4 O = 4 *16 = 64
Total = 169
1 mol = 169 grams.
0.432 mol = x
1/0.432 = 169/x
x = 0.432 * 169
x = 73.0 grams
PH + pOH = 14
11.8 + pOH = 14
pOH = 14 - 11.8
pOH = 2.2
[OH-] = 10 ^- pOH
[OH-] = 10 ^- 2.2
[OH-] = <span>6.33 x 10^-3 M
</span>
Answer B
hope this helps!
The activity series goes top to bottom, most active to least active elements, going: Li, K, Ba, Sr, Ca, Na, Mg, Mn, Zn, Fe, Cd, Co, Ni, Sn, Pb, H, Cu, Ag, Hg, Au.
Thus, your list of metals would go from most reactive to least reactive: Li, K, Mg, Zn, Fe, Cu, Au
Answer:
The Empirical Formula.
Explanation:
From the empirical formula and using the weight (in g) of a given substance, we can come up with the molecular formula which is the actual weight of a substance. Sometimes, we find that the empircal formula is the molecular formula.