What is the molality of a solution

Chemistry

1 answer:

EastWind [94]3 years ago

7 0

Answer:

The formula for molality is m = moles of solute / kilograms of solvent. In problem solving involving molality, we sometimes need to use additional formulas to get to the final answer. One formula we need to be aware of is the formula for density, which is d = m / v, where d is density, m is mass and v is volume

Explanation:

it is a measure of the concentration of a solute in a solution in terms of amount of substance in a specified amount of mass of the solvent. This contrasts with the definition of molarity which is based on a specified volume of solution.

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Violet light has a wavelength of 422 nm, how much energy is found in this photon of violet light?

pashok25 [27]

Answer:

3.025eV

Explanation:

Energy of a photon is given by

E=hν

Where h is the Planck's constant

And ν represents the frequency of the photon .

Now for violet light wavelength λ=410nm= 410x10^-9m

By using the formula ν=c/λ

where c represents speed of light I.e 3 x 10^8 m/s

Calculating energy of violet photon we have

E=hc/λ

h=6.626 x 10^-34 J.s

On inserting the value in the formula , we get

E= 4.848 x 10^-19 Joules

To convert it into electron volts we need to divide it by 1.602 x10^-19.

The final result what we have is E=3.026eV.

Note: the result calculated is according to the above value . result may vary according to the values taken .

3 0

3 years ago

How many moles of sodium hydroxide would react with 1 Mole of sulphuric acid?

Paul [167]

Answer:

Two moles.

Explanation:

Sulphuric (sulfuric) acid $\rm H_2SO_4$ is a diprotic acid. When one mole of $\rm H_2SO_4$ molecules dissolve in water, two moles of $\rm H^{+}$ ions would be produced.

$\rm H_2SO_4 \to 2\, H^{+} + {SO_4}^{2-}$ .

On the other hand, sodium hydroxide $\rm NaOH$ is a monoprotic base. When one mole of $\rm NaOH$ formula units dissolve in water, only one mole of hydroxide ions $\rm OH^{-}$ would be produced.

$\rm NaOH \to Na^{+} + OH^{-}$ .

Note that $\rm H^{+}$ and $\rm OH^{-}$ react at a one-to-one ratio:

$\rm H^{+} + OH^{-} \to H_2O$ .

As a result, it would take $2\; \rm mol$ of $\rm OH^{-}$ to react with the $\rm 2\; mol$ of $\rm H^{+}$ that was released when $1\; \rm mol$ of $\rm H_2SO_4$ is dissolved in water. Since one mole of $\rm NaOH$ formula units could produce only one mole of $\rm OH^{-}$ , it would take $\rm 2\; mol$ of $\rm NaOH$ formula units to produce that $2\; \rm mol$ of $\rm OH^{-}$ for reacting with $1\; \rm mol$ of $\rm H_2SO_4$ .