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Natali5045456 [20]

3 years ago

12

What is the approximate bond angle around the central carbon atom in acrolein?

1 answer:

Soloha48 [4]3 years ago

7 0

Answer:

The approximate bond angle around the central carbon atom in acrolein is 120°.

Explanation:

The structure of acrolein is shown in the attachment. From the structure, we can deduce that the central carbon atom is in an sp2 hybridization (Atoms with a double bond hybridize in an sp2 fashion).

Atoms with sp2 hybridization have trigonal planar geometry, in this kind of hybridization, bonds are oriented the farthest away possible from each other, to minimize overlapping and the angle that allows that is 120°.

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Answer:

72g

Explanation:

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3 years ago

2 HI(g) ⇄ H2(g) + I2(g) Kc = 0.0156 at 400ºC 0.550 moles of HI are placed in a 2.00 L container and the system is allowed to rea

Alex_Xolod [135]

<u>Answer:</u> The concentration of hydrogen gas at equilibrium is 0.0275 M

<u>Explanation:</u>

Molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}$

Moles of HI = 0.550 moles

Volume of container = 2.00 L

$\text{Initial concentration of HI}=\frac{0.550}{2}=0.275M$

For the given chemical equation:

$2HI(g)\rightleftharpoons H_2(g)+I_2(g)$

<u>Initial:</u> 0.275

<u>At eqllm:</u> 0.275-2x x x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[H_2][I_2]}{[HI]^2}$

We are given:

$K_c=0.0156$

Putting values in above expression, we get:

$0.0156=\frac{x\times x}{(0.275-2x)^2}\\\\x=-0.0458,0.0275$

Neglecting the negative value of 'x' because concentration cannot be negative

So, equilibrium concentration of hydrogen gas = x = 0.0275 M

Hence, the concentration of hydrogen gas at equilibrium is 0.0275 M

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3 years ago

Z forms chloride compounds with the formulae ZCl2 and ZC13.

nordsb [41]

Answer:

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hope this answer is helpful for u.

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2 years ago

What form of matter is made from only one type of atom? compound molecule material element

Lelechka [254]

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2 years ago

Read 2 more answers

A certain radioactive nuclide has a half life of 1.00 hour(s). Calculate the rate constant for this nuclide. s-1 Calculate the d

Karo-lina-s [1.5K]

Answer:

k= 1.925×10^-4 s^-1

1.2 ×10^20 atoms/s

Explanation:

From the information provided;

t1/2=Half life= 1.00 hour or 3600 seconds

Then;

t1/2= 0.693/k

Where k= rate constant

k= 0.693/t1/2 = 0.693/3600

k= 1.925×10^-4 s^-1

Since 1 mole of the nuclide contains 6.02×10^23 atoms

Rate of decay= rate constant × number of atoms

Rate of decay = 1.925×10^-4 s^-1 ×6.02×10^23 atoms

Rate of decay= 1.2 ×10^20 atoms/s

8 0

3 years ago