Do electrons have psrticle nature or wave nature?

How many atoms are in 3.05 grams of copper?

vfiekz [6]

6.02 x10^23 atom

3.5g x 1mol/63.55g Cu x 6.02 x 10^23/ 1mol=

3.32 x 10^22 atoms

5 0

3 years ago

A liquid has an empirical formula CCl2, and a boiling point of 1 21 oC. When vapourised, the gaseous compound has a density of 4

Darya [45]

Based on the data given, the molar mass of the gas is 165.5 g/mol while the molecular weight of the gas is 165.5 amu

<h3>How can molar mass of a gas be obtained from density, temperature and pressure?</h3>

The molar mass of a gas can be obtained from density, temperature and pressure using the formula below:

molar mass = density × molar gas constant × temperature/pressure

Molar gas constant, R = R = 0.082 L.atm/mol/K.

Temperature = 150 °C = 423 K

Pressure = 785 torr = 1.033 atm

density = 4.93 g/L

molar mass of gas = 4.93 × 0.082 × 423/1.033

molar mass of gas = 165.5 g/mol

Then, molecular weight of the gas = 165.5 amu

Therefore, the molar mass of the gas is 165.5 g/mol while the molecular weight of the gas is 165.5 amu

Learn more about molar mass of a gas at: brainly.com/question/26215522

6 0

2 years ago

Help me plz, I need help on this.

kirza4 [7]

Omg i lost everything ugh
To do it again

1. 12g+2(16g)= 44g/mol
25.01/ 44g/mol= .... mol

2. 14g+3(1g)= 17g/mol
34.05g/ 17g/mol=.... mol

3. 23g+1g+ 12g+ 3(16g)= 84g/mol
17.31g/ 84g/mol=.... mol

4. 6(12g)+12(1g)+6(16g)= 180g/mol
123.44g/ 180g/mol=.... mol

5. 23g+16g+1g= 40g/mol
2.2mol x 40g/mol= .... g

6. 2(35g)= 71g/mol
4.5mol x 71g/mol= .... g

7. 137g+ 2(14g)+ 6(16g)= 261g/mol
0.002mol x 261g/mol= ....g

8. 2(56g)+ 3(32g)+ 12(16g)= 400g/mol
5.4mol x 400g/mol=.... g

I cant believe i had to do this all over

4 0

3 years ago

Which types of electron orbitals will have higher energy than a 4d orbital?

alexdok [17]

Answer:

D) 4f

Explanation:

To determine which electron orbital that will have higher energy than a 4d orbital, we write the electron configuration starting with s-orbital.

1s

2s 2p

3s 3p 3d 3f

4s 4p 4d 4f

5s 5p 5d 5f

6s 6p 6d 6f

7s 7p 7d 7f

In ascending order, 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 3f, <u>4d</u><em>,</em> 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 6f, 7d, 7f.

From the electronic configuration formula above, the electron orbitals that have higher energy than a 4d orbital are 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 6f, 7d, 7f.

Therefore, 4f is the correct answer.

8 0

3 years ago

On the basis of the information above, what is the approximate percent ionization of HNO2 in a 1.0 M HNO2 (aq) solution?

enot [183]

Answer:

The answer is "2%"

Explanation:

Equation:

$HNO_2\ (aq) \leftrightharpoons H^{+} \ (aq) + NO_2^{-}\ (aq) \\\\\ K_a = 4.0\times \ 10^{-4}$