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1 year ago

A. what is the hybridization of the central atom in cse2?

Chemistry

1 answer:

elixir [45]1 year ago

4 0

The orbital hybridization of the central carbon atom in CSe2 is sp.

In chemical bonding, atomic orbitals may be combined to form appropriate hybrid orbitals suitable for bonding. The orbitals that combine during hybridization must be close enough in energy.

In the compound Cse2, carbon is the central atom bonded to two selenium atoms. The carbon atom in CSe2 is sp hybridized.

Learn more about orbital hybridization: brainly.com/question/1869903

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Why doesn't glue stick to the inside of the bottle? (Not sure if this is chemistry or not lol)

Leni [432]

Answer:

When white glue is in the bottle, there's not enough air inside the bottle to cause the water to evaporate to make the glue sticky. Basically, the bottle holds the glue from the air and keeps the glue from going everywhere.

8 0

2 years ago

A child has a toy balloon with a volume of 1.80 L. The temperature of the balloon when it was filled was 293 K at a pressure of

Mekhanik [1.2K]

Answer:

2.42L

Explanation:

Given parameters:

V₁ = 1.8L

T₁ = 293K

P₁ = 101.3kPa

P₂ = 67.6kPa

T₂ = 263K

Unknown:

V₂ = ?

Solution:

To solve this problem, we are going to use the combined gas law to find the final volume of the gas. The combined gas law expression combines the equation of Boyle's law, Charles's law and Avogadro's law;

$\frac{P_{1} V_{1} }{T_{1} } = \frac{P_{2} V_{2} }{T_{2} }$

All the units are in the appropriate form. We just substitute and solve for the unknown;

101.3 x 1.8 / 293 = 67.6 x V₂ / 263

V₂ = 2.42L

7 0

2 years ago

What is the Ka of a weak acid (HA) if the initial concentration of weak acid is 4.5 x 10-4 M and the pH is 6.87? (pick one)

rewona [7]

Answer:

Ka = $4.04 \times 10^{-11}$

Explanation:

Initial concentration of weak acid = $4.5 \times 10^{-4}\ M$

pH = 6.87

$pH = -log[H^+]$

$[H^+]=10^{-pH}$

$[H^+]=10^{-6.87}=1.35 \times 10^{-7}\ M$

HA dissociated as:

$HA \leftrightharpoons H^+ + A^{-}$

(0.00045 - x) x x

[HA] at equilibrium = (0.00045 - x) M

x = $1.35 \times 10^{-7}\ M$

$Ka = \frac{[H^+][A^{-}]}{[HA]}$

$Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 - 0.000000135}$

0.000000135 <<< 0.00045

$Therefore, Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 } = 4.04 \times 10^{-11}$

5 0

2 years ago

What is the mean free path for the molecules in an ideal gas when the pressure is 100 kPa and the temperature is 300 K given tha

sladkih [1.3K]

Answer:

The mean free path = 2.16*10^-6 m

Explanation:

Given:

Pressure of gas P = 100 kPa

Temperature T = 300 K

collision cross section, σ = 2.0*10^-20 m2

Boltzmann constant, k = 1.38*10^-23 J/K

To determine:

The mean free path, λ

Calculation:

The mean free path is related to the collision cross section by the following equation:

$\lambda =\frac{1}{n\sigma }------(1)$

where n = number density

$n = \frac{P}{kT}-----(2)$

Substituting for P, k and T in equation (2) gives:

$n = \frac{100,000 Pa}{1.38*10^{-23} J/K*300K} =2.42*10^{25}\ m^{3}$

Next, substituting for n and σ in equation (1) gives:

$\lambda =\frac{1}{2.42*10^{25}m^{-3}* .0*10^{-20}m^{2}}=2.1*10^{-6}m$

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2 years ago

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Slav-nsk [51]

Answer:

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6 0

1 year ago

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