PH scale is used to determine how acidic, basic or neutral a solution is
pH can be calculated using the H₃O⁺
ph can be calculated as follows
pH = - log[ H₃O⁺]
[H₃O⁺] = 1 x 10⁻⁹
pH = - log [1 x 10⁻⁹]
pH = 9
pH of solution is 9
Answer:
Explanation:
CH₃COOH + NaOH = CH₃COONa + H₂O .
42.5 mL of .115 M of NaOH will contain .0425 x .115 moles of NaOH
= 48.875 x 10⁻⁴ moles NaOH
It will react with same number of moles of acetic acid
So number of moles of acetic acid in 3.45 mL = 48.875 x 10⁻⁴
number of moles of acetic acid in 1000 mL = 48.875 x 10⁻⁴ x 10³ / 3.45 moles
= 1.4167 moles
= 1.4167 x 60 gram
= 85 grams .
So 85 grams of acetic acid will be contained in one litre of acetic acid.
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Answer:
38.16 mL
Solution:
Data Given;
Density = 1.31 g/mL
Mass = 50 g
Volume = ?
Formula Used;
Density = Mass ÷ Volume
Solving for Volume,
Volume = Mass ÷ Density
Putting values,
Volume = 50 g ÷ 1.31 g.mL⁻¹
Volume = 38.16 mL
Answer:
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