<span>An unsaturated solution is one that has a concentration lower than it's solubility. What happens is all of the solute dissolves in the solvent. For example, dumping a spoon of sugar into coffee ends up with an unsaturated solution. Because all the sugar dissolves into the solvent (the coffee).</span>
I think the answer would be Ga3+
Answer:
Explanation:
Fossil fuel is nonrenewable, whereas hydro, wind, solar, and wave power are generally ... depletes its fuel stock but at a relatively slow rate, and geothermal energy (Williamson, ... Future primary energy sources (e.g. nuclear, coal and solar) are primarily ... b. Distributed peak electrical generation, •. c. Distributed electricity ...
Answer:
The answers are in the explanation.
Explanation:
The energy required to convert 10g of ice at -10°C to water vapor at 120°C is obtained per stages as follows:
Increasing temperature of ice from -10°C - 0°C:
Q = S*ΔT*m
Q is energy, S specific heat of ice = 2.06J/g°C, ΔT is change in temperature = 0°C - -10°C = 10°C and m is mass of ice = 10g
Q = 2.06J/g°C*10°C*10g
Q = 206J
Change from solid to liquid:
The heat of fusion of water is 333.55J/g. That means 1g of ice requires 333.55J to be converted in liquid. 10g requires:
Q = 333.55J/g*10g
Q = 3335.5J
Increasing temperature of liquid water from 0°C - 100°C:
Q = S*ΔT*m
Q is energy, S specific heat of ice = 4.18J/g°C, ΔT is change in temperature = 100°C - 0°C = 100°C and m is mass of water = 10g
Q = 4.18J/g°C*100°C*10g
Q = 4180J
Change from liquid to gas:
The heat of vaporization of water is 2260J/g. That means 1g of liquid water requires 2260J to be converted in gas. 10g requires:
Q = 2260J/g*10g
Q = 22600J
Increasing temperature of gas water from 100°C - 120°C:
Q = S*ΔT*m
Q is energy, S specific heat of gaseous water = 1.87J/g°C, ΔT is change in temperature = 20°C and m is mass of water = 10g
Q = 1.87J/g°C*20°C*10g
Q = 374J
Total Energy:
206J + 3335.5 J + 4180J + 22600J + 374J =
30695.5J =
30.7kJ
Answer:
5.6 seconds
Explanation:
The reaction follows a zero-order in dinitrogen monoxide
Rate = k[N20]^0 = change in concentration/time
[N20]^0 = 1
Time = change in concentration of N2O/k
Initial number of moles of N2O = 300 mmol = 300/1000 = 0.3 mol
Initial concentration = moles/volume = 0.3/4 = 0.075
Number of moles after t seconds = 150 mmol = 150/1000 = 0.15 mol
Concentration after t seconds = 0.15/4 = 0.0375 M
Change in concentration of N2O = 0.075 - 0.0375 = 0.0375 M
k = 0.0067 M/s
Time = 0.0375/0.0067 = 5.6 s