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saul85 [17]
3 years ago
12

Identify each definition that applies to the compound in red. Check all that apply. HCI + NaOH → H2O + NaCl

Chemistry
2 answers:
Xelga [282]3 years ago
5 0

Answer:

A. Arrhenius acid    

B. Bronsted-Lowry acid

Explanation:

kenny6666 [7]3 years ago
4 0

Answer:

Arrhenius acid      

Bronsted-Lowry acid

Explanation: just answered on e2020

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Pls answer with a passion and a goal of smartness
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Answer:

b

Explanation:

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10. How much protein one should obtain from diet<br> (1 gram, 3 gram,4gram)
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How much protein one should obtain from diet

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4 gram

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What is enthalpy?
aleksandrvk [35]

Answer: the heat content of a system at constant pressure.

Explanation:

Enthalpy is defined as the heat content of a system at constant pressure.

It is the heat absorbed or released during a reaction at constant pressure,denoted as ΔH.

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3 years ago
Radiation fog forms at night when the skies are clear and the relative humidity
LenKa [72]

Answer: A. True

Explanation:

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3 years ago
n an experiment, 39.26 mL of 0.1062 M NaOH solution was required to titrate 37.54 mL of \ v unknown acetic acid solution to a ph
olasank [31]

Answer:

Molarity: 0.111M

% (w/w): 0.666

Explanation:

The reaction of NaOH with acetic acid (CH₃COOH) is:

NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O

<em>where 1 mole of NaOH reacts per mole of acetic acid producing 1 mole of water and 1 mole of sodium acetate.</em>

As 39.26mL ≡ 0.03926L of 0.1062M are required to titrate the solution of acetic acid. Moles are:

0.03926L × (0.1062mol / L) = 4.169x10⁻³ moles of NaOH. As 1 mole of NaOH reacts per mole of acetic acid:

4.169x10⁻³ moles of CH₃COOH.

Molarity is defined as ratio between moles of substance and volume of solution in liters. Thus, molarity of acetic acid solution is:

4.169x10⁻³ moles of CH₃COOH / 0.03754L = <em>0.111M</em>

<em></em>

As molar mass of acetic acid is 60g/mol, 4.169x10⁻³ moles weights:

4.169x10⁻³ moles × (60g / mol) = <em>0.2501 g of acetic acid</em>

Now, assuming density of solution as 1.00g/mL, 37.54mL weights <em>37.54g</em>.

Thus, percent by weight is:

0.2501g CH₃COOH / 37.54g × 100 = <em>0.666% (w/w)</em>

8 0
2 years ago
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