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3 years ago

What is the bond angle of a methane molecule that contains four hydrogen atoms bonded to one carbon atom?

2 answers:

6 0

Methane is a tetrahedral molecule where each C-H is kept as far away from the others as possible due to electron-electron repulsion, resulting in a bond angle of 109.5˚ (Technically 109˚ 28' but I would just use the decimal version if I were you).
Hope this helps!

Marta_Voda [28]3 years ago

5 0

Answer:

Explanation:on plato

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What is the name of this molecule?

Neko [114]

The triple bond is in the first place, not the 3rd because we have to choose the smallest possible number. That means the name will be: ‘but-1-yne’ or ‘1-butyne’

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3 years ago

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The partial pressure of CO2 gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO2 gas (in g) will be released f

frutty [35]

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. We can calculate the concentration of CO₂ using Henry's law.

$C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 4.60 atm = 7.59 \times 10^{-3} M$

We can calculate the mass of CO₂ in 1.1 L considering its molar mass is 44.01 g/mol.

$\frac{7.59 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.367 g$

Now, we will repeat the same procedure for a partial pressure of 1.28 atm.

$C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 1.28 atm = 2.11 \times 10^{-3} M$

$\frac{2.11 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.102 g$

The mass of CO₂ released will be equal to the difference in the masses at the different pressures.

$m = 0.367 g - 0.102 g = 0.265 g$

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

Learn more: brainly.com/question/18987224

<em>The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO₂ gas (in g) will be released from 1.1 L of the carbonated water when the partial pressure of CO2 is lowered to 1.28 atm? At 25 ºC, the Henry’s law constant for CO₂ dissolved in water is 1.65 x 10⁻³ M/atm, and the density of water is 1.0 g/cm³.</em>

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2 years ago

Write the following number using Scientific notation : 13,215,296.50 *

ludmilkaskok [199]

Answer:

13.22x10^7

Explanation:

7 0

3 years ago

Will give lots of points if answered correctly. Determine the kb for chloroform when 0.793 moles of solute in 0.758 kg changes t

Liono4ka [1.6K]

Answer: The value of $K_{b}$ for chloroform is $3.62^{o}C/m$ when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

Explanation:

Given: Moles of solute = 0.793 mol

Mass of solvent = 0.758

$\Delta T_{b} = 3.80^{o}C$

As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.

$Molality = \frac{no. of moles}{mass of solvent (in kg)}\\= \frac{0.793 mol}{0.758 kg}\\= 1.05 m$

Now, the values of $K_b$ is calculated as follows.

$\Delta T_{b} = i\times K_{b} \times m$

where,

i = Van't Hoff factor = 1 (for chloroform)

m = molality

$K_{b}$ = molal boiling point elevation constant

Substitute the values into above formula as follows.

$\Delta T_{b} = i\times K_{b} \times m\\3.80^{o}C = 1 \times K_{b} \times 1.05 m\\K_{b} = 3.62^{o}C/m$

Thus, we can conclude that the value of $K_{b}$ for chloroform is $3.62^{o}C/m$ when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

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3 years ago

During a chemical reaction, the______

WINSTONCH [101]

Answer:

limiting reagent :)

Explanation:

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2 years ago

Read 2 more answers