The best answer for this is A) I’m pretty sure it is
ΔS = 0.250 J·K^(-1)
<em>Step 1</em>. Calculate the <em>moles of CO</em>
From the<em> </em><em>Ideal Gas Law</em>,
<em>n</em> = (<em>pV</em>)/(<em>RT</em>) = (8.00 kPa × 3.00 L)/(8.314 kPa·L·K^(-1)·mol^(-1) × 298.15 K)
= <em>0.009 682 mo</em>l
Δ<em>S</em> = <em>nC</em>_Vln(<em>T</em>_2/<em>T</em>1)
= 0.0096 82 mol × 20.17 J·K^(-1)mol^(-1) ln(1073.15 K/298.15 K)
= 0.1953 J·K^(-1) × ln3.599 = 0.250 J·K^(-1)
Answer:
a theory of a experiment and the experiment acts like a judge determining if the hypothesis is right or wrong from the final test result
Explanation:
Answer:
1. d. The reaction is spontaneous in the reverse direction at all temperatures.
2. c. The reaction is spontaneous at low temperatures.
Explanation:
The spontaneity of a reaction is associated with the Gibbs free energy (ΔG). When ΔG < 0, the reaction is spontaneous. When ΔG > 0, the reaction is non-spontaneous. ΔG is related to the enthalpy (ΔH) and the entropy (ΔS) through the following expression:
ΔG = ΔH - T. ΔS [1]
where,
T is the absolute temperature (T is always positive)
<em>1. What can be said about an Endothermic reaction with a negative entropy change?</em>
If the reaction is endothermic, ΔH > 0. Let's consider ΔS < 0. According to eq. [1], ΔG is always positive. The reaction is not spontaneous in the forward direction at any temperature. This means that the reaction is spontaneous in the reverse direction at all temperatures.
<em>2. What can be said about an Exothermic reaction with a negative entropy change?</em>
If the reaction is exothermic, ΔH < 0. Let's consider ΔS < 0. According to eq. [1], ΔG will be negative when |ΔH| > |T.ΔS|, that is, at low temperatures.
The specific heat in, J /g °C of copper is 0.386J/g°C
HOW TO CALCULATE SPECIFIC HEAT CAPACITY OF A SUBSTANCE:
- The specific capacity of a substance can be calculated using the following formula:
Q = m × c × ∆T
Where;
Q = quantity of heat absorbed or released (J)
m = mass of substance (g)
∆T = change in temperature (°C)
c = specific heat capacity (J/g°C)
For a colorimeter, the following equation applies:
m.c.∆T (water) = - m.c.∆T (metal)
m = 50g
c = 4.184 J/g°C
∆T = 29.2 - 26°C = 3.2°C
m = 70g
c = ?
∆T = 29.2 - 54°C = -24.8°C
m.c.∆T (water) = - m.c.∆T (metal)
50 × 4.184 × 3.2 = -(70 × c × -24.8)
669.44 = 1736c
c = 669.44 ÷ 1736
c = 0.386J/g°C
Therefore, the specific heat in, J /g °C of copper is 0.386J/g°C
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