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yawa3891 [41]
3 years ago
10

Balance this equation. If a coefficient of "1" is required, choose "blank" for that box.

Chemistry
2 answers:
sweet [91]3 years ago
8 0
<span>C2H4 + O2 → CO2 + H2O
</span>
We want the same number of each element on each side. 
Since there is only one compound on each side, we can start with carbon: When we double C02, we end up with 2 carbon atoms on each side. 
When we double H20, we end up with 4 hydrogen atoms on each side.
The right side now contains 4 + 2 = 6 oxygen atoms, so we put a 3 in front of 02.
The result is:
C2H4 + 3O2 → 2CO2 + 2H2O
Andrej [43]3 years ago
4 0
When balancing  the  equation  you  need  to  know  how  many     moles  reacted  and  how  many were formed.To  balance  this  equation  we  require  1 moles  of  C2H4,  3moles  of  O2  to  form 2  moles CO2  and 2 moles  H2o
therefore the  equation is  represented  as
 C2H4+3O2 ->2CO2+2H2O
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Which unit can be used to express the rate of a reaction?
Ugo [173]
<span>A) mL / s

This is the amount of milliliters per second</span>
4 0
3 years ago
Read 2 more answers
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
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MAXImum [283]

Answer:

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Explanation:

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Answer:

MARK AS BRAINLIEST

Explanation:

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