The molarity of the lake water is 0.00001 M and the pH of lake water is 5.
The lake water is acidic.
Explanation:
Data given:
molarity of base solution Mbase = 0.1 M
volume of the base solution Vbase = 0.1 ml or 0.0001 litre
volume of lake water Vlake = 1000ml or 1 litre
molarity of the lake water, Mlake = ?
Using the formula for titration:
Mbase X Vbase = Mlake X
Mlake = 
Putting the values in the equation:
Mlake = 
Mlake = 0.00001 M
The pH of the lake water will be calculated by using the following formula:
pH = -
[
]
pH = -
[ 0.00001]
pH = 5
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Answer: 160.40 g Fe2O3 are needed.
Explanation:
Balanced equation: 2 Al + Fe2O3 —> Al2O3 + 2Fe
54.2 g Al * 1 mol Al / 26.98 g Al * 1 mol Fe2O3 / 2 mol Al * 159.69 g Fe2O3 / 1 mol = 160.40 g Fe2O3 are needed.
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