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Marianna [84]
3 years ago
7

The concentration of hydronium ions is greater than 1 × 10–7 for basic solutions. True False

Chemistry
1 answer:
Nataly [62]3 years ago
5 0

Answer : The given statement is, false.

Explanation :

Ionic product of water : It is defined as the product of the concentration of hydrogen ion and the concentration of hydroxide ion.

The mathematical expression will be,

[H^+][OH^-]=10^{-14}

When the concentration of hydrogen ion and hydroxide ion are equal then the solution is neutral and the value will be, 1\times 10^{-7}. And the pH of the solution is, 7.

When the concentration of hydrogen ion is greater than 1\times 10^{-7} then the solution is acidic in nature. And the pH of the solution is, less than 7.

When the concentration of hydrogen ion is less than 1\times 10^{-7} then the solution is basic in nature. And the pH of the solution is, greater than 7.

Hence, the given statement is false.

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2 years ago
If she suddenly dropped the books, this would increase their energy. If she instead decided to hold the books over her head, thi
Anna35 [415]

Answer:

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3 years ago
What is the molality of a solution of 12.9 g of fructose (C6H12O6) in 31.0 g of water?
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Explanation:

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8 0
3 years ago
A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a
Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = m C .ΔT

m is mass of the solution = 151.8 mL * \frac{1 g}{1 mL} = 151.8 g

C is the specific heat of solution = 4.18 \frac{J}{g. ^{0}C}

ΔT is the temperature change = 31.50^{0}C - 21.45^{0}C = 10.05^{0}C

q = 151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J

Moles of NaOH = 101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol NaOH

Moles of H_{2}SO_{4} = 50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}

Enthalpy of the reaction = \frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol

5 0
3 years ago
Calculate the Ph of:
andrey2020 [161]

Answer:

pH= 3.82

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Ethanoic acid or acetic acid (CH3COOH), ionises partially because it is a weak acid.

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3 years ago
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