Chemistry, 23.03.2021 05:20 Maria3737

Match the description with the sound feature. amplitude
the part of a wave where the particles
are closer together
frequency
the density of the medium's particles
at the compressions of the wave
rarefaction
the distance between compressions
or rarefactions
wavelength
the part of a wave where the particles
are farther apart
compression
the number of wavelengths that pass
a fixed point in one second

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Atomicity occurs to prevent updates of the database occurring only partially, which can cause greater problems than rejecting the whole series outright.some are monatomic, diatomic, triatomic etc because it shows the number of atoms. for example, monatomic means it contains only one atom.

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vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in $h_{2}s$ , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between $h_{2}s$ molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent $h_{2}s$ molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.

Answer from: Quest

so she was walking and she stepped on warm jeffy , uh uh uhh dadye i just pooped my pants

Answer from: Quest

the missing options are :

(1) lower boiling point and a lower freezing point(2) lower boiling point and a higher freezing point(3) higher boiling point and a lower freezing point(4) higher boiling point and a higher freezing point

answer : the correct answer is option (3) higher boiling point and a lower freezing point

explanation :

when a non volatile solute is added to a solvent like water, it results in elevation of boiling point.

let us find the increase in boiling point of water when cacl₂ is added to water to make a solution of 1 m

step 1 : find molality of the solution

let us assume we have 1 l of solution.

$m = \frac{moles (solute)}{l ( solution)}$

$1 m = \frac{moles (cacl_{2})}{1 l }= 1 mol cacl_{2}$

let us convert this to grams of cacl₂.

molar mass of cacl2 is 111 g/mol

grams of cacl₂ = $1 mol \times \frac{111g}{mol} = 111 g cacl_{2}$

mass of solute = 111 g

assuming density of solution same as water which is 1g/ml we have ,

mass of solution = density * volume

mass of solution = $\frac{1g}{ml} \times 1000 ml = 1000 g$

mass of solution = mass of solvent + mass of solute

mass of solute = 1000g - 111g

mass of solute 889 g = 0.889 kg

the molality of the solution can be calculated as

$molality = \frac{mol (solute)}{kg ( solvent)}$

$molality = \frac{1mol}{0.889kg}$

molality = 1.12 m

step 2 : find boiling point elevation.

the elevation in boiling point is calculated as

$\bigtriangleup t_{b} = i \times k_{b} \times m$

here δtb = boiling point elevation

i = vant hoff factor which is the number of ions formed by the solute .

cacl₂ dissociates as $cacl_{2} \rightarrow ca^{2+} + 2 cl^{-}$

it forms 3 ions . therefore,

i = 3

m = molality which is 1.12 m

kb is constant for a given solvent. for water it is 0.512

$\bigtriangleup t_{b} = 3 \times 0.512 \times 1.12$

$\bigtriangleup t_{b} = 1.7$

the boiling point of water increases by 1.7 °c

step 3: find freezing point depression.

when a non volatile solute is added to water, it results in decrease in freezing point.

the formula to find freezing point depression is

$\bigtriangleup t_{f} = i \times k_{f} \times m$

kf is constant for water and its value is 1.86

$\bigtriangleup t_{f} = 3 \times 1.86 \times 1.12$

$\bigtriangleup t_{f} = 6.2$

the freezing point of water decreases by 6.2 °c

therefore we can say that, the solution will have a higher boiling point and a lower freezing point

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