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AnnyKZ [126]
3 years ago
13

An element has an atomic weight of 137 g/mol and a BCC crystal structure. If the volume density of the element is 2.2 g/cm3 what

is the radius (in nm) of the atom?

Chemistry
1 answer:
kap26 [50]3 years ago
7 0

Explanation:

Below is an attachment containing the solution to the question.

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Why can’t there be a temperature lower than absolute zero
Firdavs [7]
Atoms cannot physically get more compact (temperature lowering increases the space between atoms) At absolute zero, the atoms are as physically close together as allowed by the physical laws governing it.
7 0
3 years ago
A radioactive nucleus emits a beta particle, then the parent and daughter nuclei are
Nana76 [90]

Answer:

isobars

Explanation:

How?

  • A radioactive nucleus emits beta particle(Like uranium,radium)
  • So the mass numbers are same for daughter nuclei .
  • They have different atomic numbes .

So they are isobars

3 0
2 years ago
Read 2 more answers
At sea level, where the pressure was 104 kPa and temperature 21.1 ºC, a certain mass of air occupies 2.0 m3 . To what volume wil
Romashka [77]

Answer:

The volume of air at where the pressure and temperature are  52 kPa, -5.0 ºC is 3.64 m^3.

Explanation:

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 104 kPa

P_2 = final pressure of gas = 52 kPa

V_1 = initial volume of gas = 2.0m^3

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 21.1^oC=273+21.1=294.1K

T_2 = final temperature of gas = -5.0^oC=273+(-5.0)=268 K

Now put all the given values in the above equation, we get:

\frac{104 kPa\times 2.0m^3}{294.1 K}=\frac{52 kPa\times V_2}{268 K}

V_2=3.64 m^3

The volume of air at where the pressure and temperature are  52 kPa, -5.0 ºC is 3.64 m^3.

3 0
3 years ago
What is the difference in mass between 3.01×10^24 atoms of gold and a gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm
Zanzabum

Answer:

The difference in mass between 3.01×10^24 atoms of gold and a gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm is :

<u>Difference</u>  <u>in mass</u> =<u> 985.32 - 984.5 = 0.82 g</u>

Explanation:

<u>Part I :</u>

n =\frac{3.01\times 10^{24}}{6.022\times 10^{23}}

n = 4.9983

n = 4.99 moles

(Note : You can also take n = 5 mole )

Molar mass of gold = 196.96 g/mole

This means, 1 mole of gold(Au) contain = 196.96 grams

So, 4.99 moles of gold contain = 5\times 196.96 g

4.99 moles of gold contain = 984.8 g

Mass of {3.01\times 10^{24}} atoms of gold = 984.5 g

<u>Part II :</u>

Density of Gold = 19.32 g/cm^{3}

Volume of the cuboid = length\times breadth\times height

Volume of the gold bar =6.00\times 4.25\times 2.00

Volume of the gold bar = 51cm^{3}

Using formula,

Density = \frac{mass}{Volume}

Mass = Density\times Volume

Mass = 19.32 \times 51

Mass = 985.32 g

So, A  gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm has mass of <u>985.32 g</u>

<u>Difference</u>  <u>in mass</u> =<u> 985.32 - 984.5 = 0.82 g</u>

3 0
3 years ago
In most cases, reactant-favored reactions have:​
vekshin1

Answer:

Have negitive values

Explanation:

4 0
3 years ago
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