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2 years ago

What is the melting point of a solution in which 3.5 grams of sodium chloride is added to 230 mL of water?

2 answers:

8 0

We are going to use this equation:

ΔT = - i m Kf

when m is the molality of a solution

i = 2

and ΔT is the change in melting point = T2- 0 °C

and Kf is cryoscopic constant = 1.86C/m

now we need to calculate the molality so we have to get the moles of NaCl first:

moles of NaCl = mass / molar mass

= 3.5 g / 58.44

= 0.0599 moles

when the density of water = 1 g / mL and the volume =230 L

∴ the mass of water = 1 g * 230 mL = 230 g = 0.23Kg

now we can get the molality = moles NaCl / Kg water

=0.0599moles/0.23Kg

= 0.26 m

∴T2-0 = - 2 * 0.26 *1.86

∴T2 = -0.967 °C

mart [117]2 years ago

5 0

Answer:

0.952 °C

Explanation:

The change in melting point is computed as:

ΔT = k*m*i

where ΔT is the difference between the melting point of water and of solution, k is a constant (1.86 °C*kg/mol for water), i is the van't Hoff factor (equal to 2 for sodium chloride because 2 ions are obtained after its dissolution), and m is the molality of the solution.

Molar mass of sodium chloride: 58.44 g/mol

Moles of of sodium chloride: mass / molar mass 3.5/58.44 = 0.059 mol

Density of water 1 kg/L

230 mL of water are equivalent to 0.23 L

mass of water: density * volume = 1*0.23 = 0.23 kg

Molality of the solution: m = moles of solute/ kg of solvent = 0.059/0.23 = 0.256

Finally:

ΔT = 1.86*0.256*2 = 0.952 °C

Water melting point: 0 °C

So, the solution melting point is: 0 - 0.952 = 0.952 °C

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It takes 839./kJmol to break a carbon-carbon triple bond. Calculate the maximum wavelength of light for which a carbon-carbon tr

tresset_1 [31]

Answer:

The maximum wavelength of light for which a carbon-carbon triple bond could be broken by absorbing a single photon is 143 nm.

Explanation:

It takes 839 kJ/mol to break a carbon-carbon triple bond.

Energy required to break 1 mole of carbon-carbon triple bond = E = 839 kJ

E = 839 kJ/mol = 839,000 J/mol

Energy required to break 1 carbon-carbon triple bond = E'

$E'=\frac{ 839,000 J/mol}{N_A}=\frac{839,000 J}{6.022\times 10^{23} mol^{-1}}=1.393\times 10^{-18} J$

The energy require to single carbon-carbon triple bond will corresponds to wavelength which is required to break the bond.

$E'=\frac{hc}{\lambda }$ (Using planks equation)

$\lambda =\frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{1.393\times 10^{-18} J}$

$\lambda =1.427\times 10^{-7} m =142.7 nm = 143 nm$

$(1 m = 10^9 nm)$

The maximum wavelength of light for which a carbon-carbon triple bond could be broken by absorbing a single photon is 143 nm.

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2 years ago

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Kaylis [27]

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2 years ago

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mr Goodwill [35]

Answer:

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2 years ago

If the lead can be extracted with 92.5% efficiency, what mass of ore is required to make a lead sphere with a 3.50 cm radius? le

ElenaW [278]

The volume of sphere can be calculated using the following formula:

$V=\frac{4}{3}\pi r^{3}$

Here, r is radius of the sphere which is 3.5 cm. Putting the value,

$V=\frac{4}{3}\pi r^{3}=\frac{4}{3}(3.14)(3.5cm)^{3}=180 cm^{3}$

This is equal to the volume of lead, density of lead is $11.34 g/cm^{3}$ thus, mass of lead can be calculated as follows:

m=d×V

Putting the values,

$m=11.34 g/cm^{3}\times 180 cm^{3}=2041.2 g$

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m=0.686×0.866=0.5940 g

Therefore, 0.5940 g of lead is obtained from 1 g of ore for 100% efficiency, thus, for 92.5% efficiency

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1 g of lead obtain from $\frac{1}{0.54945}=1.82$ grams of ore.

Thus, 2041.2 g of lead obtain from:

$2041.2\times 1.82=3.715\times 10^{3}g or 3.715 kg$

Therefore, mass of ore required to make lead sphere is 3.715 kg.

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