Please help me find the answer , thank you.

If pure water boils at 99.8 degrees celcius, what is the expected elevated boiling point of a solution of 2.50g of CaCl2, in 50.

Alexeev081 [22]

INFORMATION:

We know that:

- pure water boils at 99.8 degrees celcius

And we must calculate the expected elevated boiling point of a solution of 2.50g of CaCl2, in 50.0mL (i.e., 50.0g) of H2O

STEP BY STEP EXPLANATION:

To calculate it, we need to use that:

Boiling point of solution = boiling point of pure solvent + boiling point elevation (ΔTb)

The elevation in boiling point (ΔTb) is proportional to the concentration of the solute in the solution. It can be calculated via the following equation.

$ΔTb=i\times k_b\times m$

Where,

- i is the Van’t Hoff factor

- Kb is the ebullioscopic constant

- m is the molality of the solute

From given information, we know that:

- i = 3

Now, the ebullioscopic constant (Kb) is often expressed in terms of °C * kg * mol^-1. The value of Kb for water is 0.512.

So, kb = 0.512 °C * kg * mol^-1

Then, we must calculate the molality

$\begin{gathered} Molality=\frac{\text{ moles of solute}}{\text{ kg of solvent}} \\ Molality=\frac{\frac{2.5g}{110.98\frac{g}{mol}}}{0.05kg}=0.45\frac{mol}{kg} \end{gathered}$

So, m = 0.45 mol/kg

Replacing the values in the formula for ΔTb

$\begin{gathered} ΔT_b=3\times0.512\frac{\degree C\cdot kg}{mol}\times0.45\frac{mol}{kg} \\ ΔT_b=0.69\degree C \end{gathered}$

Finally, the expected elevated boiling point of the solution would be

$\text{ Boling point of solution}=99.8\degree C+0.69\degree C=100.49\degree C$

ANSWER:

The expected elevated boiling point of the solution is 100.49 °C

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How many photons are produced in a laser pulse of 0.897 J at 407 nm?

Leona [35]

The relationship between energy of a single photon and its wavelength can be determined using the formula E=hc/lambda where E is energy, h is Planck's constant, c is the speed of light, and lambda is photons.

Before being able to solve for energy, need to convert nanometers to meters.

407 nm (1 m/1 x 10^9 nm) = 4.07 x 10^-7 m

Then plug in the values we know into the equation.
E h(Planck's constant) c(speed of light)
E = (6.63 x 10^-34 Js)(3 x 10^8 m/s) / 4.07 x 10^-7 m (lambda)

E=(0.000000000000000000000000000000000663js)(300,000,000m/s)=1.989×10^-25j/ms

E=1.989x10^-25j/ms /{divided by} 4.07x10^-7m = 4.8869779x10^-33 J (the meters cancel out)

E = 4.89 x 10^-33 J

This gives us the energy in Joules of a single photon. Now, we can find the number of photons in 0.897 J

0.897J / 4.89 x 10^-33 J = ((0.897 J) / 4.89) x ((10^(-33)) J) = 1.8343558 x 10^-34

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How many grams of KOH are dissolved in 2.50 liters of 1.50 M KOH? (The molar mass of KOH = 56.11 g/mol) 210 grams 93.5 grams 15

romanna [79]

<span>There are a number of ways to express concentration of a solution. This includes molarity. Molarity is expressed as the number of moles of solute per volume of the solution. We can calculate as follows:

Mass of KOH = 1.50 mol KOH/ L solution (2.50 L) (56.11 g/mol) = 210.41 g KOH

Therefore, the first option is the answer. </span><span />

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