Answer : The mass of of water present in the jar is, 298.79 g
Solution : Given,
Mass of barium nitrate = 27 g
The solubility of barium nitrate at
is 9.02 gram per 100 ml of water.
As, 9.02 gram of barium nitrate present in 100 ml of water
So, 27 gram of barium nitrate present in
of water
The volume of water is 299.33 ml.
As we know that the density of water at
is 0.9982 g/ml
Now we have to calculate the mass of water.


Therefore, the mass of of water present in the jar is, 298.79 g
Answer:
T = 9.875K
Explanation:
The ideal gas Law is PV = nRT.
P = Pressure
V = Volume
n = amount of substance
R = 8.314 J/(K. mol)
T = Temperature in Kelvin
22g CO2
CO2 Molar Mass = 44g/mol
C = 12g/mol
O = 16g/mol
P = 0.8210atm
V = 50L
PV = nRT
0.8210 x 50 = 8.314 x 0.5 x T
41.05 = 4.157T
T = 41.05/4.157
T = 9.875K
3501.75
is the answer
i think
(your welcome)
Answer:
The energy profile for rotation about the C-C bond in ethane is shown in the image, along with the Newman projections of the corresponding ethane conformer.
Explanation:
If you see the ethane molecule (second image) from the C-C bond axis (third image), as in the Newman projections, it's easy to draw an angle between one of the hydrogen atoms of the visible carbon, the carbon itself, and one of the hydrogens of the hidden carbon.
When you make a rotation about the C-C bond, the angle between those hydrogens will change. If you start with an eclipsed conformation, with each hydrogen of the hidden C exactly behind the hydrogens of the visible C, the angle will be 0°, or also 120° or 240°, as this rotations will be equivalent. On the other hand, if the angle is 60° (or 180°, or 300°), you will have a staggered conformation. The eclipsed conformation is less stable than the staggered one, because the interactions between hydrogens will be bigger (the repulsion between their electrons), and because of that the eclipsed conformations will be found in the maxima, while the staggered one will be found in the minima.