Answer:
v = 46.5 m/s
Explanation:
Given data:
Mass of car = 1210 kg
Momentum of car = 56250 kg m/s
Velocity of car = ?
Solution:
Formula:
p = mv
p = momentum
m = mass
v = velocity
Now we will put values in formula:
56250 kg m/s = 1210 kg × v
v = 56250 kg m/s / 1210 kg
v = 46.5 m/s
So a car having mass of 1210 kg with momentum 56250 kg m/s having 46.5 m/s velocity.
Answer:
Explanation:
So here's the thing . Sugar molecule is a molecular solid . It a relatively weak molecule
We all know the formula of glucose ( which is a sugar type molecule ) which is C12 H22 O 11 . This is basically bonded together by dipole dipole forces ( remember that this force of attraction is weak ) .on the other hand you have the solution ( which is usually water) .water is held by strong attraction forces known as hydrogen bonding .since this type of bonding is greater than dipole dipole forces .therefore when glucose is dropped in water . water is easily able to obercome the forces that are keeping the glucose molecule intact . therefore the molecule dissolves and breaks into pieces. Hope this helps
Answer:
The protonated form is predominant when aspirin is absorbed more readily. The ratio of conjugate base to acid is 1 to 100.
Explanation:
Aspirin is more readily absorbed when it is protonated, that is when pH is lower than pKa (<em>more H⁺ available in the medium</em>). We can confirm this using Henderson-Hasselbalch equation for pH = 1.5:

When aspirin is absorbed more readily the ratio of conjugate base to acid is 1 to 100, being the acid the <em>predominant</em> form.
Answer:
Option A
Explanation:
A) Yes. The reaction reaches equilibrium when the rate of reaction of the reverse reaction is equal to the rate of the forward reaction , then the only cause for the reverse reaction to be favoured is that the initial rate of the reverse was greater than the forward one.
B) No. The rate constant of the reverse reaction can be greater than the forward one but the rate also depends on concentrations, thus a reverse reaction with greater rate constant can result in the net reaction proceeding in the forward reaction, the reverse reaction or be at equilibrium depending on the concentrations or reactants and products
C) No. A lower activation energy means a higher rate constant , but a higher rate constant does not mean that the net reaction will proceed to the reactants ( see point B)
D) No. The energy changes determine conditions under thermodynamic equilibrium and therefore the net direction of the reaction will depend on the temperature and concentrations of reactants and products with respect to the equilibrium conditions.