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3 years ago

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Chemistry

1 answer:

Marysya12 [62]3 years ago

7 0

H⁣⁣⁣⁣ere's l⁣⁣⁣ink t⁣⁣⁣o t⁣⁣⁣he a⁣⁣⁣nswer:

bit. $^{}$ ly/3a8Nt8n

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Given the different molecular weights, dipole moments, and molecular shapes, why are their molar volumes nearly the same?

Marina CMI [18]

Answer:

option d

Explanation:

Molecular sizes of gaseous molecules are very less. Volume occupied by the all the molecules of the gases are very less or negligible as compared to the container in which it is kept. Therefore, most of the volume occupied by gaseous molecules are negligible.

Volume occupied by the gaseous molecules are actually the volume of the container and its does not depend upon the amount, molecular mass or dipole moment of the gaseous molecules.

Therefore, the correct option is d ‘Because most of the volume occupied by the substance is empty space.’

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You have a cold and symptoms include a runny nose, fever and coughing. How is your body working together to get rid of the bacte

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Your white blood cells are fighting the pathogen (sickness)

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A metal ion Mⁿ⁺ has a single electron. The highest energy line in its emission spectrum occurs at a frequency of 2.961 x 10¹⁶ Hz

Whitepunk [10]

Answer:

z≅3

Atomic number is 3, So ion is Lithium ion ( $Li^+$ )

Explanation:

First of all

v=f*λ

In our case v=c

c=f*λ

λ=c/f

where:

c is the speed of light

f is the frequency

$\lambda=\frac{3*10^8}{2.961*10^{16}}\\ \lambda=1.01317*10^{-8} m$

Using Rydberg's Formula:

$\frac{1}{\lambda}=R*z^2*(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Where:

R is Rydberg constant= $1.097*10^7$

z is atomic Number

For highest Energy:

n_1=1

n_2=∞

$\frac{1}{1.01317*10^{-8}}=1.097*10^{7}*z^2*(\frac{1}{1^2}-\frac{1}{\inf})\\z^2=8.99\\z=2.99$

z≅3

Atomic number is 3, So ion is Lithium ion ( $Li^+$ )

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3 years ago

CH3OH can be synthesized by the following reaction.

puteri [66]

Answer:

A: There is 43.12 Liter of H2 needed

B: There is 21.56 liter of CO needed

Explanation:

Step 1: Data given

CO(g)+2H2(g)?CH3OH(g)

For 1 mol of CO consumed, we need 2 moles of H2 to produce 1 mol of CH3OH

Molar mass of CO = 28.01 g/mol

Molar mass of H2 = 2.02 g/mol

Molar mass of CH3OH = 32.04 g/mol

Step 2: What volume of H2 gas (in L), measured at 754mmHg and 90?C, is required to synthesize 23.0g CH3OH?

Pressure = 754 mmHg = 0.992 atm

Temperature = 90°C = 363 Kelvin

mass of CH3OH produced = 23.0 grams

Step 3: Calculate moles of CH3OH

Moles CH3OH = mass CH3OH / Molar mass CH3OH

Moles CH3OH = 23.0 grams / 32.04 g/mol

Moles CH3OH = 0.718 moles

Step 4: Calculate moles of H2

For 1 mol of CO consumed, we need 2 moles of H2 to produce 1 mol of CH3OH

For 0.718 moles CH3OH produced, we have 2*0.718 moles =1.436 moles of H2 and 0.718 moles of CO

Step 5: Calculate volume of H2

p*V = n*R*T

with p = the pressure = 0.992 atm

with V the volume = TO BE DETERMINED

with n = the number of moles = 1.436 moles H2

with R = the gasconstant = 0.08206 L*atm/ K*mol

with T = the temperature = 363 Kelvin

V = (n*R*T)/p

V = (1.436*0.08206*363)/0.992

V = 43.12 L

Step 6: Calculate volume of CO

p*V = n*R*T

with p = the pressure = 0.992 atm

with V the volume = TO BE DETERMINED

with n = the number of moles = 0.718 moles CO

with R = the gasconstant = 0.08206 L*atm/ K*mol

with T = the temperature = 363 Kelvin

V = (n*R*T)/p

V = (0.718*0.08206*363)/0.992

V = 21.56 L

A: There is 43.12 Liter of H2 needed

B: There is 21.56 liter of CO needed

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