Answer:
Yes, it is possible. Let us consider an example of two solutions, that is, solution A having 20 percent mass RbCl (rubidium chloride) and solution B is having 15 percent by mass NaCl or sodium chloride.
It is found that solution A is having more concentration in comparison to solution B in terms of mass percent. The formula for mass percent is,
% by mass = mass of solute/mass of solution * 100
Now the formula for molality is,
Molality = weight of solute/molecular weight of solute * 1000/ weight of solvent in grams
Now molality of solution A is,
m = 20/121 * 1000/80 (molecular weight of RbCl is 121 grams per mole)
m = 2.07
Now the molality of solution B is,
m = 15/58.5 * 1000/85
m = 3.02
Therefore, in terms of molality, the solution B is having greater concentration (3.02) in comparison to solution A (2.07).
Answer:
Reduction
Explanation:
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Mn⁺⁷ +3e⁻ → Mn⁴⁺
Mn gets three electrons , its oxidation state reduced from +7 to +4 so Mn gets reduced.
Examples:
Consider the following reactions.
4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl
the oxidation state of copper is changed from +2 to +1 so copper get reduced.
CO + H₂O → CO₂ + H₂
the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.
H₂S + 2NaOH → Na₂S + 2H₂O
The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.
(C) 0.1 mole of NaCl dissolved in 1,000. mL of water
<u>Explanation:</u>
The conductivity of 0.1 mole of NaCl dissolved in 1000 mL of water will be greatest as the number of ions in 0.1 mole of NaCl will be more than 0.001, 0.05 and 0.005 moles of NaCl. Greater the number of ions in the solution, greater will be the conductivity. Specific Conductivity decreases with a decrease in concentration. Since the number of ions per unit volume that carry current in a solution decrease on dilution. Hence, concentration and conductivity are directly proportional to each other.
Answer:
Option C. Energy Profile D
Explanation:
Data obtained from the question include:
Enthalpy change ΔH = 89.4 KJ/mol.
Enthalpy change (ΔH) is simply defined as the difference between the heat of product (Hp) and the heat of reactant (Hr). Mathematically, it is expressed as:
Enthalpy change (ΔH) = Heat of product (Hp) – Heat of reactant (Hr)
ΔH = Hp – Hr
Note: If the enthalpy change (ΔH) is positive, it means that the product has a higher heat content than the reactant.
If the enthalpy change (ΔH) is negative, it means that the reactant has a higher heat content than the product.
Now, considering the question given, the enthalpy change (ΔH) is 89.4 KJ/mol and it is a positive number indicating that the heat content of the product is higher than the heat content of the reactant.
Therefore, Energy Profile D satisfy the enthalpy change (ΔH) for the formation of CS2 as it indicates that the heat content of product is higher than the heat content of the reactant.
Answer:
if the heat of reaction is positive, the reaction is said to be endothermic; if negative, exothermic.
Explanation:
hope this helps
i am so sorry if this is the wrong answer
