Answer: 4.5 x 10e-7
Explanation: 450 x 1e+9 = correct answer
Multiply amount of nanometers by 1e+9 to get the approximate result in meters.
Answer: 122 moles
Procedure:
1) Convert all the units to the same unit
2) mass of a penny = 2.50 g
3) mass of the Moon = 7.35 * 10^22 kg (I had to arrage your numbers because it was wrong).
=> 7.35 * 10^22 kg * 1000 g / kg = 7.35 * 10^ 25 g.
4) find how many times the mass of a penny is contained in the mass of the Moon.
You have to divide the mass of the Moon by the mass of a penny
7.35 * 10^ 25 g / 2.50 g = 2.94 * 10^25 pennies
That means that 2.94 * 10^ 25 pennies have the mass of the Moon, which you can check by mulitiplying the mass of one penny times the number ob pennies: 2.50 g * 2.94 * 10^25 = 7.35 * 10^25.
5) Convert the number of pennies into mole unit. That is using Avogadros's number: 6.022 * 10^ 23
7.35 * 10^ 25 penny * 1 mol / (6.022 * 10^ 23 penny) = 1.22* 10^ 2 mole = 122 mol.
Answer: 122 mol
Answer:
Explanation:The final homogenous solution, after cooling it to 40°C, will contain 47 g of potassium sulfate disolved in 150 g of water, so you can calculate the amount disolved per 100 g of water in this way:
[47 g of solute / 150 g of water] * 100 g of g of water = 31.33 grams of solute in 100 g of water.
So, when you compare with the solutiblity, 15 g of solute / 100 g of water, you realize that the solution has more solute dissolved with means that it is supersaturated.
To make a saturated solution, 15 grams of potassium sulfate would dissolve in 100 g of water.
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Answer:
Step 1;
q = w = -0.52571 kJ, ΔS = 0.876 J/K
Step 2
q = 0, w = ΔU = -7.5 kJ, ΔH = -5.00574 kJ
Explanation:
The given parameters are;
= 100 N·m
= 327 K
= 90 N·m
Step 1
For isothermal expansion, we have;
ΔU = ΔH = 0
w = n·R·T·ln(
/
) = 1 × 8.314 × 600.15 × ln(90/100) = -525.71
w ≈<em> -0.52571</em> kJ
At state 1, q = w = -0.52571 kJ
ΔS = -n·R·ln(
/
) = -1 × 8.314 × ln(90/100) ≈ 0.876
ΔS ≈ 0.876 J/K
Step 2
q = 0 for adiabatic process
ΔU = 25×(27 - 327) = -7,500
w = ΔU = <em>-7.5 kJ</em>
ΔH = ΔU + n·R·ΔT
ΔH = -7,500 + 8.3142 × 300 = -5,005.74
ΔH = ΔU = <em>-5.00574 kJ</em>