Answer:
Sucrose: glucose and fructose
Explanation:
<em>What monosaccharides will result from the hydrolysis of sucrose?</em>
<em>Sucrose</em> is a <em>disaccharide</em> composed of 2 different <em>monosaccharides</em>: glucose and fructose joining by a 1 ⇒ 2 bond. These monosaccharides will be released upon the hydrolysis of sucrose.
<em>What monosaccharide will result from the hydrolysis of starch?</em>
<em>Starch</em> is a <em>polysaccharide</em> composed of numerous glucose monomers joined by glycosidic bonds (1 ⇒ 4 and 1 ⇒ 6). These monosaccharides will be released upon the hydrolysis of starch.
This doesn't need an ICE chart. Both will fully dissociate in water.
Assume HClO4 and KOH reacts with one another. All you need to do is determine how much HClO4 will remain after the reaction. Calculate pH.
Step 1:
write out balanced equation for the reaction
HClO4+KOH ⇔ KClO4 + H2O
the ratio of HClO4 to KOH is going to be 1:1. Each mole of KOH we add will fully react with 1 mole of HClO4
Step 2:
Determining the number of moles present in HClO4 and KOH
Use the molar concentration and the volume for each:
25 mL of 0.723 M HClO4
Covert volume from mL into L:
25 mL * 1L/1000mL = 0.025 L
Remember:
M = moles/L so we have 0.025 L of 0.723 moles/L HClO4
Multiply the volume in L by the molar concentration to get:
0.025L x 0.723mol/L = 0.0181 moles HClO4.
Add 66.2 mL KOH with conc.=0.273M
66.2mL*1L/1000mL = .0662 L
.0662L x 0.273mol/L = 0.0181 moles KOH
Step 3:
Determine how much HClO4 remains after reacting with the KOH.
Since both reactants fully dissociate and are used in a 1:1 ratio, we just subtract the number of moles of KOH from the number of moles of HClO4:
moles HClO4 = 0.0181; moles KOH = 0.0181, so 0.0181-0.0181 = 0
This means all of the HClO4 is used up in the reaction.
If all of the acid is fully reacted with the base, the pH will be neutral = 7.
Determine the H3O+ concentration:
pH = -log[H3O+]; [H3O+] = 10-pH = 10-7
The correct answer is 1.0x10-7.
Answer:
THE EMPIRICAL FORMULA OF THE SUBSTANCE IS C2H5NO
Explanation:
The steps involved in calculating the empirical formula of this substance in shown in the table below:
Element Carbon Hydrogen Nitrogen Oxygen
1. % Composition 40.66 8.53 23.72 27.09
2. Mole ratio =
%mass/ atomic mass 40.66/12 8.53/1 23.72/14 27.09/16
= 3.3883 8.53 1,6943 1.6931
3. Divide by smallest
value (0.6931) 3.3883/1.6931 8.53/1.6931 1.6943/1.6931 1.6931/1.6931
= 2.001 5.038 1.0007 1
4. Whole number ratio 2 5 1 1
The empirical formula = C2H5NO
The answer is C as it slowly destroys the ozone layer
