Answer:
5.167 kJ
Explanation:
We have to divide the heating process into two steps: one for the heating process of liquid water (1) and the other for the phase transition from liquid water to steam at 100°C (2)
1 - heating from 52.1°C to 100°C:
heat(1) = m x Cp x ΔT = 2.1 g x 4.184 J/g°C x (100°C-52.1°C) = 420.9 J
2 - vaporization at 100°C:
heat(2) = m x ΔHv = 2.1 g x 2260 J/g = 4746 J
Finally, we add the heat values of the steps:
heat required = heat(1) + heat(2) = 420.9 J + 4746 J = 5166.9 J
Since 1 kJ= 1000 J, we convert from J to kJ:
5166.9 J x 1 kJ/1000 J = 5.1669 kJ ≅ 5.167 kJ
Answer:
For each system listed in the first column of the table below, decide (If possible) whether the change described in the second column will Increase the entropy S of the system, decrease S, or leave S unchanged. If you don't have enough information to decide, check the not enough information" button in the last column.
System As A few grams of water vapor (H,0). The water condenses to a liquid at a not enough The carbon dioxide is heated from -5.0 °C to 13.0°c and is also compressed from a volume of 9.0 1 to a volume of 5.0L A few moles of carbon dioxide (CO2) gas. not enough information -15.0 °C to 86.0 °C while the volume is held constant at 2.0 L A few moles of carbon dioxide (CO,) gas. S > 0 not enough 1 Don't Know
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I believe this is correct (I used the complementary base)
The number 6.022 × 1023 indicating the number of atoms or molecules in a mole of any substance
Answer:
Keiko should mix 20 mL 1% solution and 80 mL 6% solution for to make 100 mL 5% solution
Explanation:
There are 2 unknown values X= mL 6% solution and Y=1% solution. So, we need 2 equations:
1. Equation acid concentration. X mL 6% + Y mL 1% = 100 mL 5%
2. Equation solvent concentration X mL 94% + Y mL 99% = 100 mL 95%
When clearing X and Y :
(X mL 6% + Y mL 1% = 100 mL 5%) (-15,7)
X mL 94% + Y mL 99% = 100 mL 95%
_______________________________
- Y 0.83 = 16.5
Y = 19.9 mL 1% solution
Replace Y in anyone equation and X = 80 mL 6% solution
I hope to see been helpful
