Answer:
E
Explanation:
Here in this question, what we will do is to select which of the pairs that do not correlate.
A. Enthalpy and heat content
This two terms are at par with each other. By definition, the enthalpy of a system simply is the total amount of heat content it has.
B. Endothermic reaction and +H
These two terms are at par with each other. An endothermic reaction is one in which heat is absorbed from the surroundings. It has a positive value for the heat content i.e the enthalpy is positive and thus H is positive.
C. Exothermic reaction and -H
An exothermic reaction is one in which heat is released to the environment. It usually has a negative value for the enthalpy and thus the value of H is negative.
D. High energy and High Stability
These two terms are not at par. When an entity has or is of high energy, it is usually unstable. An entity at a higher energy level will not be stable until it goes to a lower level of energy.
Thus higher energy level is associated with lesser stability while lower energy levels are associated with higher stability. The lesser the energy of an entity, the higher its stability.
This makes the option our answer.
Answer:
The percent yield of the reaction is 35 %
Explanation:
In the reaction, 1 mol of hydrazine reacts with 1 mol O₂ to produce 1 mol of nitrogen and 2 moles of water.
Let's verify the moles that were used in the reaction.
2.05 g . 1mol/ 32 g = 0.0640 mol
In the 100% yield, 1 mol of hydrazine produce 1 mol of N₂ so If I used 0.0640 moles of reactant, I made 0.0640 moles of products.
Let's use the Ideal Gases Law equation to find out the real moles of nitrogen, I made (real yield).
1atm . 0.550L = n . 0.082 . 295K
(1atm . 0.550L) / 0.082 . 295K = n → 0.0225 moles
Percent yield of reaction = (Real yield / Theoretical yield) . 100
(0.0225 / 0.0640) . 100 = 35%
Answer:
CH3COOH would be more concentrated
Explanation:
The higher the concentration value, the more concentrated it is.
The relationship between concentration, moles and volume is given by the equation;
Concentration = No of moles / Volume
5.0 grams of HCOOH dissolved in 189 mL of water
Number of moles = Mass / Molar mass = 5 / 46.03 = 0.1086 mol
Concentration = 0.1086 / 0.189 = 0.5746 mol/L
1.5 moles of CH3COOH dissolved in twice as much water
Volume = 2 * 189 = 378 ml = 0.378 L
Concentration = 1.5 / 0.378 = 3.9683 mol/L
Comparing both concentration values;
CH3COOH would be more concentrated
The heat lost by the metal should be equal to the heat
gained by the water. We know that the heat capacity of water is simply 4.186 J
/ g °C. Therefore:
100 g * 4.186 J / g °C * (31°C – 25.1°C) = 28.2 g * Cp *
(95.2°C - 31°C)
<span>Cp = 1.36 J / g °C</span>
