Answer:
A i. Internal energy ΔU = -4.3 J ii. Internal energy ΔU = -6.0 J B. The second system is lower in energy.
Explanation:
A. We know that the internal energy,ΔU = q + w where q = quantity of heat and w = work done on system.
1. In the above q = -7.9 J (the negative indicating heat loss by the system). w = 3.6 J (It is positive because work is done on the system). So, the internal energy for this system is ΔU₁ = q + w = -7.9J + 3.6J = -4.3 J
ii. From the question q = +1.5 J (the positive indicating heat into the system). w = -7.5 J (It is negative because work is done by the system). So, the internal energy for this system is ΔU₂ = q + w = +1.5J + (-7.5J) = +1.5J - 7.5J = - 6.0J
B. We know that ΔU = U₂ - U₁ where U₁ and U₂ are the initial and final internal energies of the system. Since for the systems above, the initial internal energies U₁ are the same, then we say U₁ = U. Let U₁ and U₂ now represent the final energies of both systems in A i and A ii above. So, we write ΔU₁ = U₁ - U and ΔU₂ = U₂ - U where ΔU₁ and ΔU₂ are the internal energy changes in A i and A ii respectively. Now from ΔU₁ = U₁ - U, U₁ = ΔU₁ + U and U₂ = ΔU₂ + U. Subtracting both equations U₁ - U₂ = ΔU₁ - ΔU₂
= -4.3J -(-6.0 J)= 1.7 J. Since U₁ - U₂ > 0 , U₂ < U₁ , so the second system's internal energy increase less and is lower in energy and is more stable.
Answer:
-68.4 kJ
Explanation:
<u>The standard enthalpy of vaporization = 23.3 kJ/mol</u>
<u>which means the energy required to vaporize 1 mole of ammonia at its boiling point (-33 °C).</u>
To calculate heat released when 50.0 g of ammonia is condensed at -33 °C.
This is the opposite of enthalpy of vaporization which means that same magnitude of heat is released.
<u>Thus, Q = -23.3 kJ/mol</u>
<u>Where negative sign signifies release of heat</u>
Given: mass of 50.0 g
Molar mass of ammonia = 17.034 g/mol
Moles of ammonia = 50.0 /17.034 moles = 2.9353 moles
Also,
1 mole of ammonia when condenses at -33 °C releases 23.3 kJ
2.9412 moles of ammonia when condenses at -33 °C releases 23.3×2.9353 kJ
<u>Thus, amount of heat released when 50 g of ammonia condensed at -33 °C= -68.4 kJ, where negative sign signifies release of heat.</u>
Q1. -------- a specific way that a quantity may increase over time
1. Exponential growth
2. Biotic
3. Trait
4. Succession
Answer:
1. Exponential growth
Explanation:
Exponential growth is a specific way that a quantity may increase over time. The instantaneous rate of change of a quantity with respect to time is proportional to the quantity itself. When growth becomes more rapid in relation to the growing total number, then it becomes exponential.
Example: Latest updates of COVID-19. For today, the number of infected people cases from COVID-19 is 190,000. By observing the graph, to reach 100,000 the number of infected patient cases, it took 3 months and the other 90,000 patients were infected in just 15 days.
Q2. ------- is the inherent inclination of a living organism towards a particular complex behavior.
1. Abiotic
2. Biotic
3. Innate behavior
Answer:
3. Innate behavior
Explanation:
Innate behavior is behavior that occurs naturally in all members of a species. These behaviors do not have to be learned or practiced.
Example: When a human baby born, he already knows how to suck mother breast to get milk. He never learned before. As he grew up, he knows how to grasp things.
Answer:
1.26 × 10^-8 M
Explanation:
We are given;
Number of moles of mercury (i) chloride as 0.000126 μmol
Volume is 100 mL
We are required to calculate the concentration of the solution.
We need to know that;
Concentration is also known as molarity is given by;
Molarity = Number of moles ÷ Volume
Number of moles = 1.26 × 10^-10 Moles
Volume = 0.01 L
Therefore;
Concentration = 1.26 × 10^-10 Moles ÷ 0.01 L
= 1.26 × 10^-8 M
Thus, the molarity of the solution is 1.26 × 10^-8 M
The equation : y=3x-5
<h3>Further explanation
</h3>
Straight-line equations are mathematical equations that are described in the plane of cartesian coordinates
General formula
y-y1 = m(x-x1)
or
y = mx + c
Where
m = straight-line gradient which is the slope of the line
x1, y1 = the Cartesian coordinate that is crossed by the line
c = constant
The formula for a gradient (m) between 2 points in a line
m = Δy / Δx
