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jarptica [38.1K]
2 years ago
10

A certain compound with a molar mass of 120.0 g/mol crystallizes with the sodium chloride (rock salt) structure. The length of a

n edge of the unit cell is 461 pm. What is the density of this compound
Chemistry
1 answer:
vlada-n [284]2 years ago
5 0

The density of the compound is 4.64 g/cm^{3}.

Explanation:

Sodium chloride crystallize as FCC face centered cubic structure. It consist of four atoms of NaCl. So the unknown compound which crytallizes like NaCl wil have volume as:

radius in fcc is calculated as

\sqrt{2a} = 4r

putting the value where a is length of the edge.

\sqrt{2X461}

= 651.9 = 4r  so r= 162.98 pm

volume is calculated by the formula:

\frac{4}{3} πr^{3}

putting the values,

\frac{4}{3} x 3.14 x (1.62 X 10^{-8})^3

= 4.25X 10 ^-24

1.7 X 10^{-23} cm^{3} = volume of the fcc cell.

mass of the substance = 4 x 120 x 1.6605 x 10^(-24) g/u) (as it contains four atoms)

                                     = 7.9x 10^{-23}

density = \frac{mass}{volume}

putting the values

density = \frac{ 7.9x10^-23 }{1.7 x 10^{-23} }

density = 4.64 g/ cm^{3}

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Room temperature is about 20 degrees Celsius. Explain how you could convert this temperature to kelvin. Use evidence from the fi
Taya2010 [7]

Answer:

293.15 K.

Explanation:

It is given that, the room temperature is 20 degrees Celsius.

We need to convert this temperature into kelvin.

The conversion from degrees Celsius to Kelvin is as follows :

T_k=T_c+273.15

We have, T_c=20^{\circ} C

So,

T_k=20+273.15\\\\T_k=293.15\ K

So, the room temperature is 293.15 kelvin.

8 0
2 years ago
A 0.530 M Ca(OH)2 solution was prepared by dissolving 36.0 grams of Ca(OH)2 in enough water. What is the total volume of the sol
Paladinen [302]

The concentration of a solution is the number of moles of solute per fixed volume of solution.

Concentration (C) = number of moles of solute (n) / volume of the solution (v)

we have to find the volume of the solution when 36.0 g of Ca(OH)₂ is added to water to make a solution of concentration 0.530 M

mass of Ca(OH)₂ added - 36.0 g

number of moles of Ca(OH)₂ - 36.0 g / 74.1 g/mol = 0.486 mol

we know the concentration of the solution prepared and the number of moles of Ca(OH)₂ added, substituting these values in the above equation, we can find the volume of the solution

C = n/v

0.530 mol/L = 0.486 mol / V

V = 0.917 L

answer is 0.917 L

6 0
2 years ago
Read 2 more answers
If one has a solution of 0.10 m silver nitrate and it is diluted by a factor of two, what is the new concentration
Angelina_Jolie [31]
Diluted by a factor of two means that we double the volume of the solution by adding an equal volume of the water.
if we diluted it by a factor of one so the new concentration = 0.1/2=0.05 M and diluted by a factor of two so, the new concentration will be 0.05/2 = 0.025 M
7 0
2 years ago
A 0.529-g sample of gas occupies 125 ml at 60. cm of hg and 25°c. what is the molar mass of the gas?
Llana [10]

<span>Let's </span>assume that the gas has ideal gas behavior. <span>
Then we can use ideal gas formula,
PV = nRT<span>

</span><span>Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span></span>⁻¹ K⁻¹) and T is temperature in Kelvin.<span>
<span>
</span>P = 60 cm Hg = 79993.4 Pa
V = </span>125  mL = 125 x 10⁻⁶ m³

n = ?

<span> R = 8.314 J mol</span>⁻¹ K⁻¹<span>
T = 25 °C = 298 K
<span>
By substitution,
</span></span>79993.4 Pa<span> x </span>125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 298 K<span>
                                          n = 4.0359 x 10</span>⁻³ mol

<span>
Hence, moles of the gas</span> = 4.0359 x 10⁻³ mol<span>

Moles = mass / molar mass

</span>Mass of the gas  = 0.529 g 

<span>Molar mass of the gas</span> = mass / number of moles<span>
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<span>                                    = </span>131.07 g mol</span>⁻¹<span>

Hence, the molar mass of the given gas is </span>131.07 g mol⁻¹

4 0
2 years ago
Which of the following would have the most kinetic energy?
cluponka [151]

Answer:

An airplane

Explanation:

An airplane because of its position .

5 0
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