Answer:
a)905,89 atm of pressure will be generated at 298K.
b)The sap of tree can rise upto 9,357.84 meters.
Explanation:
a)
Effective concentration of sap = c = 37 M
Osmotic pressure generate at 298K = 
Temperature ,T = 298 K
905,89 atm of pressure will be generated at 298K across the endodermis root membrane.
b)
Given that 1.0 at of pressure raises the volume of water upto height of 10.33 m
Then 905.89 atm of pressure will raise the height of water upto:
The sap of the tree can rise upto 9,357.84 meter.
Answer:
Part A. 1.355 mol/L
Part B. 0.100 mol
Part C. 74.0 mL
Explanation:
Part A.
The molar mass of luminol is 177.16 g/mol, so the number of moles at 18.0 g is:
n = mass/molar mass
n = 18.0/177.16
n = 0.1016 mol
The molarity is the number of moles divided by the volume (0.075 L)
C = 0.1016/0.075
C = 1.355 mol/L
Part B.
The number of moles is the molarity multiplied by the volume, so:
n = 5.00x10⁻² mol/L * 2.00 L
n = 0.100 mol
Part C.
To prepare a solution by dilution, we can use the equation
C1V1 = C2V2
Where C1 is the concentration of the initial (stock) solution, V1 is its volume necessary, C2 is the concentration of the diluted solution, and V2 is its volume.
Thus, C1 = 1.355 M, C2 = 0.05 M, V2 = 2.00 L
1.355V1 = 0.05*2
V1 = 0.074 L
V1 = 74.0 mL
Answer:
Six electrons are transferred in the formation of Al₂O₃.
Explanation:
Aluminium metal and Oxygen react to form Al₂O₃ as,
2 Al + 3/2 O₂ → Al₂O₃
Oxidation number of Al on left hand side is zero, while than on right hand side in Al₂O₃ is +3. Means it has lost 3 electrons per one atom and six electrons per two atoms. Also, the oxidation number of O at left hand side in O₂ is zero, while that in Al₂O₃ it is -2 per atom and -6 per 3 atoms.
So, two Al atoms have lost 6 electrons and 3 O atoms have gained six electrons.
Answer:
40.73 L.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 121.59 kPa/101.325 = 1.2 atm).
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 2.0 mol).
R is the general gas constant (R = 0.082 L.atm/mol.K),
T is the temperature of the gas in K (T = 25°C + 273 = 298 K).
<em>∴ V = nRT/P</em> = (2.0 mol)(0.082 L.atm/mol.K)(298 K)/(1.2 atm) = <em>40.73 L.</em>
