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DerKrebs [107]
2 years ago
9

What is the movement of air parallel to Earth's surface called?

Chemistry
1 answer:
loris [4]2 years ago
8 0

Answer:

Wind is the movement of air parallel to Earth's surface.

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The general rule of thumb is that the (smaller or larger) a substance's atoms and the (stronger or weaker) the bonds, the harder
Ganezh [65]
Answer is: <span>The general rule of thumb is that the smaller a substance's atoms and the stronger the bonds, the harder the substance will be. 
If the distance between atoms is higher, lesser will be attraction between electrons and protons of atoms, smaller distance means stronger atoms attraction.

</span>
5 0
3 years ago
3. Crystalline structural unit of barium metal is a body-centered cubic cell. The edge length of the unit cell is 5.02x10-8 cm.
tiny-mole [99]

Answer:

The Avogadro's  number is N_A     =  6.02289 *10^{23}

Explanation:

From the question we are told that

   The edge length is  L   = 5.02 * 10^{-8} \ cm= \frac{5.02 * 10^{-8} }{100}  =  5.02 * 10^{-10}

    The density of the metal is \rho =  5.30\ g/cm^3 = 5.30 * \frac{g}{cm^3}  * \frac{1*10^6}{1*10^3} = 5.30 *10^3kg/m^3

     The molar mass of  Ba is  Z  =  137.3 \ g/mol = \frac{137.3}{1000} =  0.1373 \  kg / mol

     

Generally the volume of a unit cell is  

       V =  L^3

substituting value

        V =  [5.02 *10^{-10}]^3

         V = 1.265*10^{-28}\ m^3  

From the question we are told that 68% of the unit cell is occupied by Ba atoms and that the structure is a metal which implies that the crystalline structure will be  (BCC),

The volume of barium atom is  

        V_a  =  \frac{V}{2}  * 0.68

substituting value

        V_a  =  \frac{ 1.265*10^{-28}}{2}  * 0.68

        V_a  = 4.301 *10^{-29} \ m^3

The Molar mass of barium is mathematically represented as

      Z  =  N_A V_a *  \rho

Where N_A is the Avogadro's number

 So  

      N_A     =   \frac{ Z}{ V_a *  \rho}

substituting value

     N_A     =   \frac{ 0.1373}{ 4.301*10^{-29} *  5.3*10^{3}}

     N_A     =  6.02289 *10^{23}

4 0
2 years ago
How many covalent bonds are generally formed by atoms with five valence electrons?
torisob [31]
Five five five five five
5 0
3 years ago
Read 2 more answers
R indicates round seeds that are dominant over the wrinkled variety, represented by r. For each numbered individual in the pedig
Dmitry [639]

Answer:

it is incomplete, is this te entire tree

Explanation:

8 0
2 years ago
A student dissolves of urea in of a solvent with a density of . The student notices that the volume of the solvent does not chan
Dimas [21]

The question incomplete , the complete question is:

A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.

Answer:

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

Explanation:

Moles of urea = \frac{18.0 g}{60 g/mol}=0.3 mol

Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)

Molarity(M)=\frac{\text{Moles of compound}}{\text{Volume of solution in L}}

Molarity of the urea solution ;

M=\frac{0.3 mol}{0.200 L}=1.50 M

Mass of solvent = m

Volume of solvent = V = 200.0 mL

Density of the urea = d = 0.95 g/mL

m=d\times V=0.95 g/mL\times 200.0 mL=190 g

m = 190 g = 190 \times 0.001 kg = 0.19 kg

(1 g = 0.001 kg)

Molality of the urea solution ;

Molality(m)=\frac{\text{Moles of compound}}{\text{Mass of solvent in kg}}

m=\frac{0.3 mol}{0.19 kg}=1.58 m

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

7 0
3 years ago
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