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Licemer1 [7]
3 years ago
7

Which variable is tested during a lab investigation

Chemistry
1 answer:
Licemer1 [7]3 years ago
5 0
Actually, it could indeed be the independent variable.
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Explanation:

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If molecules move from one place to another inside of a cell, which type of communication is this?
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Hello there.

<span>If molecules move from one place to another inside of a cell, which type of communication is this?

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8 0
3 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
How do substances and mixtures diifer
vlabodo [156]

Answer:

hello hope this helps :)

a pure substance consists only of one element or one compound. a mixture consists of two or more different substances, not chemically joined together.

Explanation:

5 0
3 years ago
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