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Margaret [11]
3 years ago
5

The metallic compound reduced to the metal by electrolysis is _____

Chemistry
1 answer:
Mice21 [21]3 years ago
3 0

Answer:

Magnesium oxide

Explanation:

When magnesium reacts with oxygen, the magnesium atoms donate electrons to O2 molecules and thereby reduce the oxygen. Magnesium therefore acts as a reducing agent in this reaction. The O2 molecules, on the other hand, gain electrons from magnesium atoms and thereby oxidize the magnesium.

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Which of the following would you expect to form an ionic bond when combined?
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Ionic bond forms between metal and non-metal.

Na is a metal, and F is a nonmetal.

Na and F form ionic bond.

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Your bedroom air conditioner blows very cold air at night but only cool air during the day. Your bedroom gets lots of direct sun
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What is the experiment

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Use the Henderson-Hasselbalch equation and your knowledge of ionization to help you answer this question. Aspirin is a weak acid
Leya [2.2K]

Answer:

The protonated form is predominant when aspirin is absorbed more readily. The ratio of conjugate base to acid is 1 to 100.

Explanation:

Aspirin is more readily absorbed when it is protonated, that is when pH is lower than pKa (<em>more H⁺ available in the medium</em>). We can confirm this using Henderson-Hasselbalch equation for pH = 1.5:

pH = pKa +log(\frac{conjugate base}{weak acid} )\\1.5=3.5 +log(\frac{conjugate base}{weak acid} )\\-2.0=log(\frac{conjugate base}{weak acid} )\\10^{-2.0} =\frac{conjugate base}{weak acid}\\\frac{conjugate base}{weak acid}=0.01=\frac{1}{100}

When aspirin is absorbed more readily the ratio of conjugate base to acid is 1 to 100, being the acid the <em>predominant</em> form.

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3 years ago
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Read 2 more answers
A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of
Georgia [21]

NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa 

NaOH + CH3COOH → CH3COONa + H2O 

Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH 

Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH 

These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L 

Molarity of CH3COOH = 0.0106/0.071 = 0.1493M 

CH3COONa = 0.0076 / 0.071 = 0.1070M 

pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74. 

pH using Henderson - Hasselbalch equation: 

pH = pKa + log ([salt]/[acid]) 

pH = 4.74 + log ( 0.1070/0.1493) 

pH = 4.74 + log 0.717 

pH = 4.74 + (-0.14) 

pH = 4.60.

7 0
3 years ago
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