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chubhunter [2.5K]
3 years ago
10

FREÉ Points out!!!!!!!!!!!​

Chemistry
2 answers:
just olya [345]3 years ago
4 0

Answer:

Explanation:

okkkkkkkkkkk

Luda [366]3 years ago
3 0

Answer:

thanks

Explanation:

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Both "Ode to the West wind" and "Ode for Melancholy"
Rama09 [41]
The answer that would best complete the given statement above would be option A. Both "Ode to the West wind" and "Ode for Melancholy" praise <span>something non-human. Hope this answers your question. Have a great day ahead!</span>
5 0
3 years ago
A sample of copper absorbs 4.31E+1 kJ of heat, resulting in a temperature rise of 6.71E+1 °C. Determine the mass (in kg) of the
olga2289 [7]

Answer: 1.67 kg

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

Q=m\times c\times \Delta T

Q = Heat absorbed=4.31\times 10^1kJ = 43100J   (1kJ=1000J)

m= mass of substance = ?

c = specific heat capacity = 0.385J/g^0C

Change in temperature ,\Delta T=T_f-T_i=6.71\times 10^1^0C=67.1^0C

Putting in the values, we get:

43100J=m\times 0.385J/g^0C\times 67.1^0C

m=1670g=1.67kg   (1kg=1000g)

Thus the mass (in kg) of the copper sample is 1.67

3 0
3 years ago
if a solution is saturated, how does the rate of dissolution of a solute compare with it's rate of crystallization?
LenKa [72]

Answer:

The same

Explanation:

In a saturated solution, the rate of dissolution is equal and the same to the rate of crystallization.

  • A saturated solution of as substance (solute) at a particular temperature is one which contains the maximum amount of the substance that can dissolve at that temperature  in the presence of the crystals of the substance.
  • It is an equilibrium system in which a solid substance is in equilibrium with its own ions in solution.
  • Therefore the rate of dissolution will the same with that of crystallization.
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
The chemical equation below shows the reaction of glucose and oxygen.
goldfiish [28.3K]
• Aerobic respiration, photosynthesis


It is aerobic because it requires oxygen to be carried out. Anaerobic on the other hand does not require oxygen.

It’s two products would be the reactants for photosynthesis because in photosynthesis the reverse reaction takes place as glucose is produced along with water from carbon dioxide (CO2) and water (H2O) in the presence of chlorophyll and sunlight.
7 0
3 years ago
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