Answer:
Monosaccharides are the simplest form of carbohydrates that cannot be hydrolyzed to smaller compounds. Monosaccharides are the basic units of carbohydrates and are also known as simple sugars.
The monosaccharides are classified on the basis of number of carbon atoms present.
Triose is a type of monosaccharide molecule, which is composed of 3 carbon atoms.
Tetrose is a type of monosaccharide molecule, which is composed of 4 carbon atoms.
Pentose is a type of monosaccharide molecule, which is composed of 5 carbon atoms.
Hexose is a type of monosaccharide molecule, which is composed of 6 carbon atoms.
D-glucose is a hexose sugar and it is the <u>most abundant monosaccharide</u> in the nature.
Answer:
g. H₂SO₄ + Ca(OH)₂ → CaSO₄ + 2H₂O.
Explanation:
Sulfuric acid is represented as H₂SO₄ and calcium hydroxide is represented as Ca(OH)₂.
- To balance a chemical reaction. we should apply the law of conservation of mass that the no. of different atoms in both sides (reactants and products).
- The only choice that represent the reactants correctly and apply the law of conversation of mass is:
<em>g. H₂SO₄ + Ca(OH)₂ → CaSO₄ + 2H₂O.</em>
<em></em>
<em>f. H₂SO₄ + 2Ca(OH)₂ → 2CaSO₄ + 3H₂O </em>
is not correct that the no. of S, and O is not equal in both reaction sides.
S in reactants (1) while in products (2).
<em>O in reactants (8) while in products (11).</em>
<em></em>
<em>h. HSO₄ + CaOH → CaSO₄ + H₂O</em>
The reactants is represented in a wrong way.
<em> j. H₂SO₄+ Ca(OH)₂ → CaSO₄ + H₂O</em>
is not correct that the no. of H and O is not equal in both reaction sides.
H in reactants (4) while in products (2).
O in reactants (6) while in products (5).
<em></em>
<em>So, the right choice is: g. H₂SO₄ + Ca(OH)₂ → CaSO₄ + 2H₂O.</em>
<em></em>
Answer:
The answer to your question is below
Explanation:
Element Atomic number Protons Electrons
S (sulphur) 16 16 16
V (vanadium) 23 23 23
B (Boron) 5 5 5
Answer:
EMPIRICAL FORMULA:
C₂H₃O₂
Explanation:
Mass of compound = 4.647 g
Mass of CO₂ = 8.635 g
Mass of H₂O = 1.767 g
Empirical formula = ?
Solution:
Percentage of C = 8.635/ 4.647 × 12/44 ×100
Percentage of C = 1.86× 12/44 ×100 = 50.7
Percentage of H = 1.767/4.647 × 2/ 18 × 100
Percentage of H = 0.38 × 2/ 18 × 100
Percentage of H = 4.18
Percentage of O = 100 - (50.7+4.18)
Percentage of O = 100 - 54.88
Percentage of O = 45.12
Number of grams atom:
Number of grams atom of C = 50.7/ 12 = 4.23
Number of grams atom of H = 4.18 /1 = 4.18
Number of grams atom of O = 45.12/ 16 = 2.82
Atomic ratio:
C : H : O
4.23/2.82 : 4.18/2.82 : 2.82/2.82
1 : 1.5 : 1
C : H : O (1 : 1.5 : 1)
C : H : O 2(1 : 1.5 : 1)
C : H : O (2 : 3 : 2)
EMPIRICAL FORMULA:
C₂H₃O₂