The Gibbs free energy change (ΔG∘) for the reaction can be estimated using the equation ΔG∘ = ΔH∘ - TΔS∘. At a temperature of 718K, if ΔG∘ is negative, the reaction will be spontaneous.
The Gibbs free energy change (ΔG∘) for a reaction can be estimated using the equation ΔG∘ = ΔH∘ - TΔS∘, where ΔH∘ is the change in enthalpy and ΔS∘ is the change in entropy. At a temperature of 718K, you can estimate ΔG∘ by substituting the given values into the equation. If the value of ΔG∘ is negative, the reaction will be spontaneous. If it is positive or zero, the reaction will not be spontaneous.
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B. Rh
C. Hf
D. Co
Answer: C
Explanation:
The amount left after 1 half life = 16*1/2 = 8 g Then after each half life the amount of x will be 4 , 2 ,1 , 1/2, 1/4 grams
That's a total of 6 half-lives. Answer
Age s rock = 6*220,000 = 1,320,000 years Answer
Answer:
3.6124 m/kg
Explanation:
Molality is calculated as moles of solute (mol) divided by kilogram of solvent (kg). Here, we can find these numbers by using the 35.4%, which gives us 35.4 g of H3PO4 and 100 g of solution to work with.
To go from grams to moles for the phosphoric acid, you need to find the molar mass of the compound or element and divide the grams of the compound or element by that molar mass.
Here, the molar mass for phosphoric acid is 97.9952 g/mol. The equation would look like this:
35.4 g x 1 mol / 97.9952 g = 0.3612422 mol
Next, the 100 g of solvent can easily be converted to 0.1 kg of solvent.
To find the molality, divide the moles of solute and kilograms of solution.
0.3612422 mol / 0.1 kg = 3.6124 m/kg
b. Some of the vapor initially present will condense.
c. The pressure in the container will be 100. mm Hg.
d. Only octane vapor will be present.
e. Liquid octane will be present.
Answer:
the final pressure (108.03 mmHg ) inside the container at 339 K is more than the vapor pressure of liquid octane (100 mmHg) at 339 K.
Hence,
b. Some of the vapor initially present will condense.
e. Liquid octane will be present.
Explanation:
Given that;
The vapor pressure of liquid octane, C8H18, is 100 mm Hg at 339 K
Initial volume of the container, V1 = 537 mL
Initial vapor pressure, P1 = 68.0 mmHg
Final volume of the container, V2 = 338 mL
Let us say that the final vapor pressure = P2
From Boyle's law,
P2V2 = P1V1
P2 * 338 = 68.0 * 537
338P2 = 36516
P2 = 36516 / 338
P2 = 108.03 mmHg
Thus, the final pressure (108.03 mmHg ) inside the container at 339 K is more than the vapor pressure of liquid octane (100 mmHg) at 339 K.
Hence,
b. Some of the vapor initially present will condense.
e. Liquid octane will be present.
Answer:
Limiting reagent: AgNO3
grams AgCl : 2.44 g AgCl
grams of excess reagent remain: 0.62 g BaCl2
Explanation:
1. Change grams to mol:
AgNO3:
5.738g x (1mol/169.87g) = 0.034 mol AgNO3
BaCl2:
4.115g x (1 mol/208.23g) = 0.020 mol BaCl2
2. Limiting reagent:
AgNO3:
0.034 mol AgNO3 x (1 mol BaCL2/ 2mol AgNO3) = 0.017 mol BaCl2
BaCl2:
0.020 mol BaCl2 x (2 mol AgNO3/1 mol BaCl2) = 0.04 mol AgNO3
Limiting reagent: AgNO3
3. Grams of AgCl produced:
Using the limiting reagent:
0.017 mol AgNO3 x (2mol AgCl / 2 mol AgNO3) = 0.017 mol AgCl
4. Change mol to grams:
0.017 mol AgCl x ( 143.32 g AgCl /1mol AgCl) =2.44 g AgCl
5. Grams of the excess reagent:
0.034 mol AgNO3 x (1 mol BaCl2 / 2 mol AgNO3) = 0.017 mol BaCl2
0.020 mol BaCl2 - 0.017 mol BaCl2 = 0.003 mol BaCl2
0.003 mol BaCl2 x ( 208.23 g BaCl2 / 1 mol BaCl2) = 0.62 g BaCl2
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