Answer:
There are other details missing in the question. i.e Assume that x is much larger than the separation d between the charges in the dipole, so that the approximate expression for the electric field along the dipole axis E = p/2πε0y3 can be used, where p is the dipole moment, and y is the distance between ions. A) What is magnitude______N B) Direction? +x-direction or -x-direction C) Is this force attractive or repulsive?
A) Magnitude of electric force = 6.576 x 10 raised to power -13 N
B) Since the force direction is always dependent on the electric field and electric field = F/q, since the chlorine has a negative charge as such the direction of the electric force will be in the X - direction
C) Since the charges are of different nature, as such the force between them will be ATTRACTIVE.
Explanation:
The detailed steps is shown in the attachment
Water could be made to boil lower than its normal boiling point of 100 degrees Celsius at 92 degrees Celsius by lowering the atmosphere or external pressure.
The boiling point is the temperature at which the vapor pressure of liquid equals the external or atmosphere pressure.
So, if you decrease the external pressure the temperature needed for the liquid reach the lower external pressure is also lower..
You can accomplish by taking the water to higher levels in the Earth (atmosphere pressure at high altitudes is lower than at sea level) or by creating vaccum.
Answer:
not goin to be rude but I don't even get the question??
Answer:
a) 88.48%
b) 0.05625 mol
Explanation:
2CH₃CH₂OH(l) → CH₃CH₂OCH₂CH₃(l) + H₂O(g) Reaction 1
CH₃CH₂OH(l) → CH₂═CH₂(g) + H₂O(g) Reaction 2
a) CH₃CH₂OH = 46.0684 g/mol
CH₃CH₂OCH₂CH₃ = 74.12 g/mol
1 mol CH₃CH₂OH ______ 46.0684 g
x ______ 50.0 g
x = 1.085 mol CH₃CH₂OH
1 mol CH₃CH₂OCH₂CH₃ ______ 74.12 g g
y ______ 35.9 g
y = 0.48 mol CH₃CH₂OCH₂CH₃
100% yield _____ 0.5425 mol CH₃CH₂OCH₂CH₃
w _____ 0.48 mol CH₃CH₂OCH₂CH₃
w = 88.48%
b) Only 0.96 mol of ethanol reacted to form diethyl ether. This means that 0.125 mol of ethanol did not react. 45% of 0.125 mol reacted to form ethylene. Therefore, 0.05625 mol of ethanol reacted by the side reaction (reaction 2). Since 1 mol of ethanol leads to 1 mol of ethylene, 0.05625 mol of ethanol produces 0.05625 mol of ethylene.
