NEED HELP ASAP, how many molecules are in 4nh3

7.7 cm<br> 9.8 cm<br> 0.00<br> 0.162 m

eimsori [14]

Answer:

Volume = 1222.5cm³

Explanation:

<em>If the question is about the volume of the rectangle:</em>

The volume of a rectangle is obtained by the multiplication of its 3 dimensions: Length, width, height.

In the problem, the length of the rectangle is 0.162m = 16.2cm

The width is 7.7cm

And the height is 9.8cm

The volume is:

Volume = 16.2cm*7.7cm*9.8cm

<h3>Volume = 1222.5cm³</h3>

5 0

3 years ago

F a shell can hold a maximum of 32 electrons, what is the value of n? 2 3 4 5

Arte-miy333 [17]

Answer: 4

Explanation:

Principle Quantum Numbers: This quantum number describes the size of the orbital. It is represented by n.

Azimuthal Quantum Number: This quantum number describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...

Magnetic Quantum Number: This quantum number describes the orientation of the orbitals. It is represented as $m_l$ . The value of this quantum number ranges from $(-l\text{ to }+l)$ . When l = 2, the value of $m_l$ will be -2, -1, 0, +1, +2.

Given : a f subshell, thus l = 3 , Thus the subshells present would be 3, 2, 1, 0 and thus n will have a value of 4.

Also electrons give are 32.

The formula for number of electrons is $2n^2$ .

$2n^2=32$

$n=4$

Thus principal quantum no will be n= 4.

8 0

3 years ago

Read 2 more answers

125 ml of nitrogen gas is collected at 70.0 degrees Celsius. The pressure

dybincka [34]

Answer: Volume of the gas at STP is 22.53 L.

Explanation:

Given : Volume = 125 mL (as 1 mL = 0.001 L) = 0.125 L

Temperature = $70^{o}C = (70 + 273) K = 343 K$

Pressure = $125 kPa = 125 kPa \times \frac{0.01 atm}{1 kPa} = 1.25 atm$

According to the ideal gas equation, the volume of given nitrogen gas is calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

$1.25 atm \times V = 1 mol \times 0.0821 L atm/mol K \times 343 K\\V = \frac{1 mol \times 0.0821 L atm/mol K \times 343 K}{1.25 atm}\\= \frac{28.1603}{1.25} L\\= 22.53 L$