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2 years ago

Number 4 carbon molecule has what hybridization orbitals ?

Chemistry

1 answer:

Alona [7]2 years ago

5 0

Answer:

The four valence atomic orbitals from an isolated carbon atom all hybridize when the carbon bonds in a molecule like CH4 with four regions of electron density. This creates four equivalent sp3 hybridized orbitals.

Explanation:

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Franklin was performing an experiment by combining hydrochloric acid and sodium hydroxide. He measured the mass of his reactant

VARVARA [1.3K]

Answer:

Explanation:

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1 year ago

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A 3.5 kg iron shovel is left outside through the winter. The shovel, now orange with rust, is rediscovered in the spring. Its ma

Anton [14]

Answer:

Explanation:

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2 years ago

At 800 K, the equilibrium constant, Kp, for the following reaction is 3.2 × 10–7. 2 H2S(g) ⇌ 2 H2(g) + S2(g) A reaction vessel a

djverab [1.8K]

Answer:

0.008945 atm

Explanation:

In the reaction:

2H2S(g) ⇌ 2 H2(g) + S2(g)

Kp is defined as:

$Kp = P_{H_{2}}^2P_{S_{2}} / P_{H_{2}S}^2$

<em>Where P is the pressure of each compound in equilibrium.</em>

If initial pressure of H2S is 3.00atm, concentrations in equilibrium are:

H2S = 3.00 atm - 2X

H2 = 2X

S2: = X

Replacing:

$3.2x10^{-7} = (2X)^2X / (3-2X)^2$

$3.2x10^{-7} = 4X^3 / 9- 6X+4X^2$

0 = 4X³ - 1.28x10⁻⁶X² + 1.92x10⁻⁶X - 2.88x10⁻⁶

Solving for X:

X = 0.008945 atm

As in equilibrium, pressure of S2 is X, <em>pressure is 0.008945 atm</em>

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2 years ago

If a 1.00 mL sample of the reaction mixture for the equilibrium constant experiment required 32.40 mL of 0.258 M NaOH to titrate

andrey2020 [161]

Answer:

The concentration of acetic acid is 8.36 M

Explanation:

Step 1: Data given

Volume of acetic acid = 1.00 mL = 0.001 L

Volume of NaOH = 32.40 mL = 0.03240 L

Molarity of NaOH = 0.258 M

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate the concentration of the acetic acid

b*Ca*Va = a*Cb*Vb

⇒with b = the coefficient of NaOH = 1

⇒with Ca = the concentration of CH3COOH = TO BE DETERMINED

⇒with Va = the volume of CH3COOH = 1.00 mL = 0.001L

⇒with a = the coefficient of CH3COOH = 1

⇒with Cb = the concentration of NaOH = 0.258 M

⇒with Vb = the volume of NaOH = 32.40 mL = 0.03240 L

Ca * 0.001 L = 0.258 * 0.03240

Ca = 8.36 M

The concentration of acetic acid is 8.36 M

6 0

2 years ago

A gas has a mass of 3.82 g and occupies a volume of 0.854 L. The temperature in the laboratory is 302 K, and the air pressure is

4vir4ik [10]

m = given mass of gas = 3.82 g

M = molar mass of gas = ?

T = temperature of laboratory = 302 K

P = air pressure = 1.04 atm = 1.04 x 101325 pa

V = volume of gas = 0.854 L = 0.854 x 10⁻³ m³

using the ideal gas equation

PV = (m/M) RT

inserting the above values

(1.04 x 101325) (0.854 x 10⁻³) = (3.82/M) (8.314) (302)

M = 106.6 g

hence the molar mass of the gas comes out to be 106.6 g

3 0

2 years ago