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.
Answer:
Mw = 179.845 g/mol
Explanation:
∴ w = 26.2 g
∴ 1 mol = 6.02 E23 molecules.......Avogadro's number
⇒N° moles = 8.77 E22 molecules * ( mol / 6.02 E23 molecules ) = 0.146 mol
⇒ Mw = 26.2 g / 0.146 mol = 179.845 g/mol
Answer:- 544.5 mL of water need to be added.
Solution:- It is a dilution problem. The equation used for solving this type of problems is:
where, 
is initial molarity and 
is the molarity after dilution. Similarly, 
is the volume before dilution and 
is the volume after dilution.
Let's plug in the values in the equation:
Volume of water added = 907.5mL - 363mL = 544.5 mL
So, 544.5 mL of water are need to be added to the original solution for dilution.
Answer:
Option D. KBr < KCl < NaCl
Explanation:
We'll begin by calculating the number of mole of each sample.
This can be obtained as follow:
For NaCl:
Mass = 1 g
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mole of NaCl =?
Mole = mass /Molar mass
Mole of NaCl = 1/58.5
Mole of NaCl = 0.0171 mole
For Kbr:
Mass = 1 g
Molar mass of KBr = 39 + 80 = 119 g/mol
Mole of KBr =?
Mole = mass /Molar mass
Mole of KBr = 1/119
Mole of KBr = 0.0084 mole
For KCl:
Mass = 1 g
Molar mass of KCl = 39 + 35.5 = 74.5 g/mol
Mole of KCl =?
Mole = mass /Molar mass
Mole of KCl = 1/74.5
Mole of KCl = 0.0134 mole
Summary
Sample >>>>>>>> Number of mole
NaCl >>>>>>>>>> 0.0171
KBr >>>>>>>>>>> 0.0084
KCl >>>>>>>>>>> 0.0134
Arranging the number of mole of the sampl in increasing order, we have:
KBr < KCl < NaCl
Answer:
New volume V2 = 92.7 Liter (Approx)
Explanation:
Given:
V1 = 106 l
T1 = 45 + 273.15 = 318.15 K
P1 = 740 mm
T2 = 20 + 273.15 = 293.15 K
P2 = 780 mm
Find:
New volume V2
Computation:
P1V1 / T1 = P2V2 / T2
(740)(106) / (318.15) = (780)(V2) / (293.15)
New volume V2 = 92.7 Liter (Approx)
