If the change in Gibbs free energy for a process is positive, the corresponding change in entropy for the universe will be positive.
<h3>What is Gibbs free energy?</h3>
This is defined as the energy used by a substance involved in a chemical reaction.
The Gibbs free energy and entropy has a direct relationship which is why a positive gibbs free energy will result in a corresponding positive entropy.
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<h3>
Answer:</h3>
0.387 J/g°C
<h3>
Explanation:</h3>
- To calculate the amount of heat absorbed or released by a substance we need to know its mass, change in temperature and its specific heat capacity.
- Then to get quantity of heat absorbed or lost we multiply mass by specific heat capacity and change in temperature.
- That is, Q = mcΔT
in our question we are given;
Mass of copper, m as 95.4 g
Initial temperature = 25 °C
Final temperature = 48 °C
Thus, change in temperature, ΔT = 23°C
Quantity of heat absorbed, Q as 849 J
We are required to calculate the specific heat capacity of copper
Rearranging the formula we get
c = Q ÷ mΔT
Therefore,
Specific heat capacity, c = 849 J ÷ (95.4 g × 23°C)
= 0.3869 J/g°C
= 0.387 J/g°C
Therefore, the specific heat capacity of copper is 0.387 J/g°C
Answer:
I'm pretty sure its solid
Maalox is the trade name for an antacid and antigas medication used for relief of heartburn, bloating, and acid indigestion in which
4 ml contains
= 320mg of aluminum hydroxide
= 320mg of magnesium hydroxide
= 32mg of simethicone
recommended doses = 4 times * 2 tea spoon = 8 tea spoon/ day
given = 1 tea spoon = 5 ml
8 tea spoon = 40 ml
hence,
amount of aluminum hydroxide = 320/4 * 40 = 3200mg = 3.2 g
amount of magnesium hydroxide = 320/4 * 40 = 3.2 g
amount of simethicone = 32/4 * 40 = 320 mg = 0.32g
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Answer:
1.7 × 10 ^42
Explanation:
Using Nernst equation
E°cell = RT/nF Inq
at equilibrium
Q=K
E°cell = 0.0257 /n Ink= 0.0592/n log K
Fe2+(aq)+2e−→Fe(s) E∘= −0.45 V
Ag+aq)+e−→Ag(s) E∘= 0.80 V
Fe(s)+2Ag+(aq)→Fe2+(aq)+2Ag(s)
balance the reaction
Fe → Fe²⁺ + 2e⁻ reversing for oxidation E° = 0.45 v
2 Ag⁺ +2e⁻ → 2Ag
n = 2 moles and K = equilibrium constant
E° cell = 0.80 + 0.45 = 1.25 V
E° cell = (0.0592 / n) log K
substitute the value into the equations and solve for K
(1.25 × 2) / 0.0592 = log K
42.23 = log K
k = 10^ 42.23
K = 1.7 × 10 ^42
