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Yanka [14]
2 years ago
14

Select the correct answer

Chemistry
1 answer:
adoni [48]2 years ago
3 0

Answer:

may be A.................. ....

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What is the hybridization of the carbon atoms in 1,3-butadiene, h2c=ch-ch=ch2?
Natalija [7]
The main formula is as follow  is explained in the attached file  (please look at the examples)

the 1,3- butadiene is h2c=ch-ch=ch2, so we have 

sp²             sp²     sp²               sp²
h2c    =     ch   -     ch         =      ch2

<span>the hybridization of the carbon atoms  is </span>sp²  : trigonal planar

4 0
3 years ago
48 When an atom of the unstable isotope Na-24 decays, it becomes an atom of Mg-24 because the Na-24 atom spontaneously releases(
Crank
(1) a beta particle is your answer. Na-24 decays through beta decay, turning a neutron into a proton, electron (beta particle), and an neutrino.
4 0
2 years ago
Consider the following reaction:
iren [92.7K]

Answer:

A. ΔG° = 132.5 kJ

B. ΔG° = 13.69 kJ

C. ΔG° = -58.59 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.

ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)

where,

n: moles

ΔH°f: standard enthalpy of formation

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))

ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)

ΔH° = 178.3 kJ

We can calculate the standard entropy of the reaction (ΔS°) using the following expression.

ΔS° = ∑np . S°p - ∑nr . S°r

where,

S: standard entropy

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))

ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)

ΔS° = 160.6 J/K. = 0.1606 kJ/K.

We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.

ΔG° = ΔH° - T.ΔS°

where,

T: absolute temperature

<h3>A. 285 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ

<h3>B. 1025 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ

<h3>C. 1475 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ

5 0
3 years ago
I don’t understand this question and need help!!
Dafna1 [17]
Medium about 3 second? Not sure lol just need more points honestly lol
6 0
2 years ago
At a consent tempture the pressure on a 2.5 l ballon is increased from 2.4 atm to 5.8 atm. what is the new volume?
marissa [1.9K]

Answer:

1.034 L

Explanation:

P1 V1 = P2 V2

P1 V1 / P2 = V2

2.4 (2.5) / 5.8 = V2 = 1.034 L

8 0
1 year ago
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