Answer:
the correct answer is Blue
Density=mass/ volume so you solve for volume and get 461.96 mL
Answer:
18 g
Explanation:
We'll begin by converting 500 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
500 mL = 500 mL × 1 L / 1000 mL
500 mL = 0.5 L
Next, we shall determine the number of mole of the glucose, C₆H₁₂O₆ in the solution. This can be obtained as follow:
Volume = 0.5 L
Molarity = 0.2 M
Mole of C₆H₁₂O₆ =?
Molarity = mole / Volume
0.2 = Mole of C₆H₁₂O₆ / 0.5
Cross multiply
Mole of C₆H₁₂O₆ = 0.2 × 0.5
Mole of C₆H₁₂O₆ = 0.1 mole
Finally, we shall determine the mass of 0.1 mole of C₆H₁₂O₆. This can be obtained as follow:
Mole of C₆H₁₂O₆ = 0.1 mole
Molar mass of C₆H₁₂O₆ = (12×6) + (1×12) + (16×6)
= 72 + 12 + 96
= 180 g/mol
Mass of C₆H₁₂O₆ =?
Mass = mole × molar mass
Mass of C₆H₁₂O₆ = 0.1 × 180
Mass of C₆H₁₂O₆ = 18 g
Thus, 18 g of glucose, C₆H₁₂O₆ is needed to prepare the solution.
<span>Answer is </span>(3)
- Sodium Nitrate.<span>
</span>Normally ionic bonds can be seen between
metals and non-metals while covalent
bonds present between
non-metals. Another thing that determines the bond nature is electronegativity
value of the atoms.
If the electronegativity difference
is high, then that bond tends to be an ionic bond.<span>
</span><span>Sodium nitrate consists of </span>Na⁺<span> and </span>NO₃⁻ ions. Hence, the bond
between Na⁺ and NO₃⁻<span> is an </span>ionic
bond. <span><span>
NO</span>₃⁻ </span><span>is made from </span>N <span>and </span>O<span>. Both are </span>non-metallic
atoms. <span>The </span>electronegativities <span>of </span>N <span>and </span>O <span>are </span>3.0 <span>and </span>3.5 <span>respectively. Hence, there is </span>no
big difference between
electronegativity values (3.5 - 3.0 = 0.5<span>). Hence, the bond
between N and O is a </span><span>covalent
bond. </span>
The empirical formula of c12h24o12 is a carbohydrate.
<h3>Carbohydrate</h3>
A carbohydrate is a biomolecule made up of carbon (C), hydrogen (H), and oxygen (O) atoms, often with a hydrogen-oxygen atom ratio of 2:1 (as in water), and so having the empirical formula Cm(H2O)n (where m may or may not be different from n). All molecules that meet this exact stoichiometric criterion are not, however, automatically categorized as being carbohydrates.
The term is most frequently used in biochemistry, where it is used as a synonym for saccharide, a class of compounds that includes sugars, starches, and cellulose. The four chemical categories of saccharides are monosaccharides, disaccharides, oligosaccharides, and polysaccharides. The smallest carbohydrates, monosaccharides and disaccharides, are sometimes referred to as sugars.
Learn more about carbohydrate here:
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