Answer:
2.48 mol/L.
Explanation:
- The molarity of the solution can be expressed as <em>the number of moles of solute in 1.0 liter of the solution, </em>(M = n / V).
- It is also can be calculated from the relation:
<em>M = (mass / molar mass) solute x (1000 / V of solution)</em>
The solute is toluene and the solvent is benzene.
mass of toluene (solute) = 57.1 g,
molar mass of toluene (solute) = 92.14 g/mol.
volume of the solution = 250 ml.
∴ M = (mass / molar mass) solute x (1000 / V of solution) = [(57.1 g / 92.14 g/mol) x (1000 / 250 ml)] = 2.48 mol/L.
Answser
i didnt do dis my mom has a answer key book
Explanation:
One electron in an atom experiences the entire positive charge of the nucleus. Coulomb's law can be used in this situation to determine the effective nuclear charge.
In contrast, the outside electrons in an atom with many electrons are drawn to the positive nucleus and repelled by the negatively charged electrons at the same time. The force between two stationary, electrically charged particles can be measured using Coulomb's law inverse-square law, also known as Coulomb's law. Conventionally, the electric force between two charged objects at rest is referred to as the Coulomb force or electrostatic force.
The electron is a subatomic particle with the symbol e or with an electric charge of one elementarily negative charge. The lepton particle family's first generation includes electrons.
Learn more about Coulomb's law here
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Of what?
Attach a link or pic
Rate law for the given 2nd order reaction is:
Rate = k[a]2
Given data:
rate constant k = 0.150 m-1s-1
initial concentration, [a] = 0.250 M
reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s
To determine:
Concentration at time t = 300 s i.e. ![[a]_{t}](https://tex.z-dn.net/?f=%5Ba%5D_%7Bt%7D)
Calculations:
The second order rate equation is:
![1/[a]_{t} = kt +1/[a]](https://tex.z-dn.net/?f=1%2F%5Ba%5D_%7Bt%7D%20%3D%20kt%20%2B1%2F%5Ba%5D)
substituting for k,t and [a] we get:
1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M
1/[a]t = 49 M-1
[a]t = 1/49 M-1 = 0.0204 M
Hence the concentration of 'a' after t = 5min is 0.020 M
