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2 years ago

WILL GIVE BRAINLIEST!!!

2 answers:

7 0

Hey there!:

ΔTf = Kf * m

Molar mass glucose = 180 g/mol

number of moles glucose:

n = mass of solute / molar mass

n = 21.5 / 180

n = 0.119 moles glucose

Amount of solvent in kg = 255/1000 = 0.255 Kg

Molality = number of moles / solvent

m = 0.119 / 0.255

m = 0.466 moles/kg

Kf for water = - 1.86 ºC/*m

Therefore:

ΔTf = Kf * m

ΔTf = (-1,86) * 0.466

ΔTf = -0.86676 ºC

hope this helps!

max2010maxim [7]2 years ago

7 0

ΔTf = Kf * m Molar mass glucose = 180 g/molthe number of moles glucose:n = mass of solute / molar massn = 21.5 / 180n = 0.119 moles glucoseAmount of solvent in kg = 255/1000 = 0.255 Kg Molality = number of moles / solventm = 0.119 / 0.255 m = 0.466 moles/kgKf for water = - 1.86 ºC/*mTherefore:ΔTf = Kf * m ΔTf = (-1,86) * 0.466ΔTf = -0.86676 ºC

this is to the best of my brain power and i hope this helps with your problem

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Assuming that all infrared vibrations are active, what is

777dan777 [17]

Answer:

The number of lines possible for SO2 is 3

Explanation:

The following Procedure should be followed when calculating the number of vibrational modes:-

Identify if the given molecule is either linear or non-linear
Calculate the number of atoms present in your molecule
Place the value of n in the formula and solve.

SO2 is a non-linear molecule because it contains a lone pair which causes the molecule to bent in shape hence, The mathematical formula for calculating the number of non-linear molecule in a infrared region is (3n - 6) here n is the number of atoms in molecule.

hence for Sulphur Dioxide (SO2), n will be 3

Therefore, The number of lines possible for SO2 is (3*3) - 6 = 3

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2 years ago

What is the smallest unit that makes up matter? A. molecules B. chemicals C. atoms D. alloys

Kamila [148]

Hello there!

An Atom would be considered to be the smallest unit that would make up matter. If were to ever see a picture of an atom, they would consists of a proton, neutron, electron, and a nucleus. And all this would be called a Atom.

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2 years ago

Read 2 more answers

Which of the following shows the 3 main jobs of a cell?

skad [1K]

C is the answer hope this helps

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2 years ago

Calculate the pH of the following simple solutions:

IceJOKER [234]

Answer :

(1) pH = 1.27

(2) pH = 13.35

(3) The given solution is not a buffer.

Explanation :

(1) 53.1 mM HCl

Concentration of HCl = $53.1mM=53.1\times 10^{-3}M$

As HCl is a strong acid. So, it dissociates completely to give hydrogen ion and chloride ion.

So, Concentration of hydrogen ion= $53.1\times 10^{-3}M$

pH : It is defined as the negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$pH=-\log (53.1\times 10^{-3})$

$pH=1.27$

(2) 0.223 M KOH

Concentration of KOH = 0.223 M

As KOH is a strong base. So, it dissociates completely to give hydroxide ion and potassium ion.

So, Concentration of hydroxide ion= 0.223 M

Now we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (0.223)$

$pOH=0.65$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-0.65=13.35$

(3) 53.1 mM HCl + 0.223 M KOH

Buffer : It is defined as a solution that maintain the pH of the solution by adding the small amount of acid or a base.

It is not a buffer because HCl is a strong acid and KOH is a strong base. Both dissociates completely.

As we know that the pH of strong acid and strong base solution is always 7.

So, the given solution is not a buffer.

5 0

2 years ago

Naphthalene, C10H8, melts at 80.2Â°C. If the vapour pressure of the liquid is 1.3 kPa at 85.8Â°C and 5.3 kPa at 119.3Â°C, use th

sweet-ann [11.9K]

(a) One form of the Clausius-Clapeyron equation is

ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂); where in this case:

P₁ = 1.3 kPa

P₂ = 5.3 kPa

T₁ = 85.8°C = 358.96 K

T₂ = 119.3°C = 392.46 K

Solving for ΔHv:

ΔHv = R * ln(P₂/P₁) / (1/T₁ - 1/T₂)

ΔHv = 8.31 J/molK * ln(5.3/1.3) / (1/358.96 - 1/392.46)

ΔHv = 49111.12 J/molK

(b) Normal boiling point means that P = 1 atm = 101.325 kPa. We use the same formula, using the same values for P₁ and T₁, and replacing P₂ with atmosferic pressure, solving for T₂:

ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂)

1/T₂ = 1/T₁ - [ ln(P₂/P₁) / (ΔHv/R) ]

1/T₂ = 1/358.96 K - [ ln(101.325/1.3) / (49111.12/8.31) ]

1/T₂ = 2.049 * 10⁻³ K⁻¹

T₂ = 488.1 K = 214.94 °C

(c) The enthalpy of vaporization was calculated in part (a), and it does not vary depending on temperature, meaning that at the boiling point the enthalpy of vaporization ΔHv is still 49111.12 J/molK.

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2 years ago