The answer is 0784 the second one
Answer:
The properties <u>that </u><u>do not vary</u><u> with the variation in the quantity of the material are called as </u><u>intensive property</u>.
While the <u>extensive properties</u><u> are those which </u><u>vary with the variation in the quantity of the material</u>.
Intensive properties:
Mass and volume
Extensive properties:
Density and melting point
Explanation:
The properties <u>that </u><u>do not vary</u><u> with the variation in the quantity of the material are called as </u><u>intensive property</u>.
While the <u>extensive properties</u><u> are those which </u><u>vary with the variation in the quantity of the material</u>.
Intensive properties:
Mass and volume, as with increase in quantity of the material mass and volume increases.
Extensive properties:
Density and melting point, as they remain constant for a object or material
Answer:
Metals and non-metals are substances that can be differentiated on the basis of various physical and chemical properties. Metals are the elements that are generally hard as strong metallic bond exists between the atoms. As against non-metals are the elements which are usually soft.
Answer:
24e⁻ are transferred by the reaction of respiration.
Explanation:
C₆H₁₂O₆ + 6O₂ → 6 H₂O + 6CO₂
This is the reaction for the respiration process.
In this redox, oxygen acts with 0 in the oxidation state on the reactant side, and -2 in the product side - REDUCTION
Carbon acts with 0 in the glucose (cause it is neutral), on the reactant side and it has +4, on the product side - OXIDATION
6C → 6C⁴⁺ + 24e⁻
In reactant side we have a neutral carbon, so as in the product side we have a carbon with +4, it had to lose 4e⁻ to get oxidized, but we have 6 carbons, so finally carbon has lost 24 e⁻
6O⁻² + 6O₂ + 24e⁻ → 6O₂²⁻ + 6O⁻²
In reactant side, we have 6 oxygen from the glucose (oxidation state of -2) and the diatomic molecule, with no charge (ground state), so in the product side, we have the oxygen from the dioxide with -2 and the oxygen from the water, also with -2 at the oxidation state. Finally the global charge for the product side is -36, and in reactant side is -12, so it has to win 24 e⁻ (those that were released by the C) to be reduced.
<span>M(NO</span>₃<span>)</span>₂<span> fully separates into M</span>²⁺<span> and NO</span>₃<span> </span>²⁻<span> </span><span>and M(OH)</span>₂<span> partially separates
as <span>M</span></span>²⁺<span><span> and 2OH</span></span>⁻
<span>M(NO</span>₃<span>)</span>₂<span><span> </span>→
M</span>²⁺<span> + 2NO</span>₃²⁻
<span>0.202 M 0.202 M</span>
<span> M(OH)</span>₂<span>(s) ↔ <span>M</span></span>²⁺<span><span> (aq) + 2OH</span></span>⁻<span><span>(aq)</span></span>
<span>I - -</span>
<span>C -X +X +2X</span>
<span>E X 2X</span>
<span>Ksp = [M</span>²⁺<span> (aq)] [OH</span>⁻<span>(aq)]</span>²
4.45 * 10∧-12 = (0.202
+ X ) (2X)²
Since X is very small, (0.202 + X ) = 0.202
<span>4.45 * 10<span>-12 </span>= 0.202 *
4X</span>²
<span> X = 2.347 </span>× 10∧-6 M
Hence
the solubility of <span>M(OH)2
is 2.347 </span>× 10∧-6 M