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2 years ago

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How many grams of water can be cooled from 41 ∘c to 19 ∘c by the evaporation of 62 g of water? (the heat of vaporization of wate

r in this temperature range is 2.4 kj/g. the specific heat of water is 4.18 j/g⋅k.)?

1 answer:

Vsevolod [243]2 years ago

8 0

62 g of water are vaporized and the energy required is 2.4 kJ/g

So 62g x 2.4 kJ/g = 148.8 kJ or 148,800 Joules

Q = mCΔT
Q is energy in joules, m is mass of water, C is the specific heat, delta T is change in temp

148,800 = m(4.18)(41 - 19) = 1618g or 1.6 kg of water

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Answer:

Explanation:

At the cathode

In case of molten AgI

Silver will be collected

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lithium will be collected

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hydrogen gas will be collected as reduction potential of H⁺ is more than Li⁺

in case of aqueous AgI,

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At the Anode

In case of molten NaBr

Bromine will be collected

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Fluorine will be collected

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2 years ago

Determine the mass of oxygen in a 7.20 g sample of Al2(SO4)3.

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Given:

7.20 g sample of Al2(SO4)3

Required:

Mass of oxygen

Solution:

Since you are not given a chemical reaction, just base your solution to the chemical formula given.

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The density of phosphorus vapor at 310 degrees celcius and 775 mmHg is 2.64g/L. what is the molecular formula of the phosphorus

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Answer:

The molecular formula of the phosphorus is P4

Explanation:

<u>Step 1:</u> Data given

Density of phosphorus vapor at 310 °C and 775 mmHg = 2.64g /L

<u>Step 2: </u>Calculate the molecular weight

We assume phosphorus to be an ideal gas

So p*V = n*R*T

⇒ with p = the pressure of phosphorus = 775 mmHg

⇒ with V = the Volume

⇒ with n = the number of moles = mass/molecular weight

⇒ with R = ideal gas constant = 0.08206 L*atm/K*mol

⇒ with T = the absolute temperature

p*V = m/MW *R*T

MW = mRT/PV

⇒ Since the volume is unknown but can be written as density = mass/volume

MW = dRT/P

MW = (2.64g/L * 0.08206 L*atm/K*mol * 583 Kelvin)/1.0197 atm

MW = 123.86 g/mol

<u>Step 3</u>: Calculate molecular formula of phosphorus

The relative atomic mass of phosphorus = 30.97 u

123.86 / 30.97 = 4

The molecular formula of the phosphorus is P4

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2 years ago

Which phrase describes the molecular structure and properties of two solid forms of carbon, diamond and graphite?(1) the same mo

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I think the answer is (4) different molecular structures and different properties. Both structure are crystal. The structure of graphite is organized in layers. And structure of diamond is a strong network of atoms. Also due to the different structures, they have different properties.

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2 years ago

The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma

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To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

$A=6R^{2}\sqrt{3}$

Here, R is radius and is related to a as follows:

$R=\frac{a}{2}$

Putting the value in expression for area,

$A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}$

The value of a is 0.4961 nm

Since, $1 nm=10^{-7}cm$

Thus, $0.4961 nm=4.961\times 10^{-8} cm$

Putting the value,

$Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}$

Now, volume can be calculated as follows:

$V=Area\times c$

The value of c is 1.360 nm or $1.360\times 10^{-7} cm$

Putting the value,

$V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}$

now, number of atom in unit cell can be calculated by using the following formula:

$n=\frac{\rho N_{A}V_{c}}{A}$

Here, A is atomic mass of $Cr_{2}O_{3}$ is 151.99 g/mol.

Putting all the values,

$n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18$

Thus, there will be 18 $Cr_{2}O_{3}$ units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of $Cr^{3+}$ and $O^{2-}$ is 62 pm and 140 pm respectively.

Converting them into cm:

$1 pm=10^{-10}cm$

Thus,

$r_{Cr^{3+}}=6.2\times 10^{-9}cm$

and,

$r_{O^{2-}}=1.4\times 10^{-8}cm$

Volume of sphere will be sum of volume of total number of cations and anions thus,

$V_{S}=V_{Cr^{3+}}+V_{O^{2-}}$

Since, volume of sphere is $V=\frac{4}{3}\pi r^{3}$ ,

$V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )$

Putting the values,

$V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}$

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

$packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758$

Thus, atomic packing factor is 0.758.

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2 years ago

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