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:
6. Color Change.
Production of an odor.
Change of Temperature.
Explanation:
That is the only one I know :)
Answer:
A. 30cm³
Explanation:
Based on the chemical reaction:
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
<em>1 mol of calcium carbonate reacts with 2 moles of HCl to produce 1 mol of CO₂</em>
<em />
To solve this question we must convert the mass of each reactant to moles. With the moles we can find limiting reactant and the moles of CO₂ produced. Using PV = nRT we can find the volume of the gas:
<em>Moles CaCO₃ -Molar mass: 100.09g/mol-</em>
1.00g * (1mol / 100.09g) = 9.991x10⁻³ moles
<em>Moles HCl:</em>
50cm³ = 0.0500dm³ * (0.05 mol / dm³) = 2.5x10⁻³ moles
For a complete reaction of 2.5x10⁻³ moles HCl there are necessaries:
2.5x10⁻³ moles HCl * (1mol CaCO₃ / 2mol HCl) = 1.25x10⁻³ moles CaCO₃. As there are 9.991x10⁻³ moles, HCl is limiting reactant.
The moles produced of CO₂ are:
2.5x10⁻³ moles HCl * (1mol CO₂ / 2mol HCl) = 1.25x10⁻³ moles CO₂
Using PV = nRT
<em>Where P is pressure = 1atm assuming STP</em>
<em>V volume in L</em>
<em>n moles = 1.25x10⁻³ moles CO₂</em>
<em>R gas constant = 0.082atmL/molK</em>
<em>T = 273.15K at STP</em>
<em />
V = nRT / P
1.25x10⁻³ moles * 0.082atmL/molK*273.15K / 1atm = V
0.028L = V
28cm³ = V
As 28cm³ ≈ 30cm³
Right option is:
<h3>A. 30cm³</h3>
By the second law of thermodynamics:
Heat can not spontaneously flow from cold regions to hot regions without external work being performed on a system.
Heat transfer is the passage of thermal energy from a hot ( t B = 80° C ) to a colder body ( t A = 40° C ).
Answer: B ) Heat flows from object B to object A.
