Answer:
The answer is D. Hydroelectric
This follows the law of conservation of momentum. Momentum is the product of mass and velocity of object.
Momentum = mass(m) x velocity(v)
law of conservation of momentum means that the total momentum of system before the collision of 2 objects is equal to the total momentum after the collision
Before the collision total momentum
= m1v1 + m2v2
m1 = 2 kg
v1 = 2 m/s
m2 = 6 kg
v2 = 0 m/s
substituting the values in the equation
total momentum before = (2 kg x 2 m/s) + (6 kg x 0 m/s)
total momentum = 4 kgm/s
after the collision the 2 objects stick together and have a common velocity
total momentum after the collision = (6 kg + 2 kg)x V = 8V
V = speed of the conglomerate particle
since total momentum before is equal to total momentum after
8V = 4
V = 2 m/s
speed of conglomerate particle is 2 m/s
Mass is density times volume
0.760 x 531 = 403.56g
404g if 3 sig figs
From the calculation, the molar mass of the solution is 141 g/mol.
<h3>What is the molar mass?</h3>
We know that;
ΔT = K m i
K = the freezing constant
m = molality of the solution
i = the Van't Hoft factor
The molality of the solution is obtained from;
m = ΔT/K i
m = 3.89/5.12 * 1
m = 0.76 m
Now;
0.76 = 26.7 /MM/0.250
0.76 = 26.7 /0.250MM
0.76 * 0.250MM = 26.7
MM= 26.7/0.76 * 0.250
MM = 141 g/mol
Learn more about molar mass:brainly.com/question/12127540?
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Answer:
See explanation below
Explanation:
You forgot to put the picture to do so. In this case, I manage to find one, and I hope is the one you are looking for. If not, then post it again and I'll gladly help you out again.
According to the picture with the answer, we have a cyclohexane with 4 methyl groups there. Two of them are facing towards the molecule with a darker bond. This means that the alkyl bromide, should have a bromine in one of the bonds, and in order to produce an E2 reaction, this bromine should be facing in the opposite direction of the methyl groups which are facing towards. This is because an E2 reaction occurs with the less steric hindrance in the molecule. If the bromine is in the same direction as the methyl group, it will cause a lot more of work to do a reaction, and therefore, an E2 reaction. I will promote instead a E1 or a sustitution product.
Therefore the alkyl bromide should be like the one in the picture 2.
