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umka2103 [35]
3 years ago
10

Which of the following best describes a hypothesis?

Chemistry
2 answers:
Serhud [2]3 years ago
6 0
It is most definitely C, A hypothesis is a prediction before an experiment occurs
lions [1.4K]3 years ago
4 0
It’s C because hypothesis is what u think would happen in the experiment
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As the mass of a sample increases, the number of moles present in the sample
vivado [14]

Answer:

increases

Explanation:

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How can balancing equations support the law of conservation
amm1812

Answer:

detail is given below.

Explanation:

This law was given by French chemist  Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

Law of conservation of mass:

According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

For example:

In given photosynthesis reaction:

6CO₂ + 6H₂O  → C₆H₁₂O₆ + 6O₂

The given equation is balanced chemical equation of photosynthesis. There are six carbon atoms, eighteen oxygen atoms and twelve hydrogen atoms on the both side of equation so this reaction followed the law of conservation of mass.

If equation is not balanced,

CO₂ + H₂O  → C₆H₁₂O₆ + O₂

It can not follow the law of conservation of mass because mass is not equal on both side of equation.

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3 years ago
5. The bond in a molecule of Cl2 could be described as
jenyasd209 [6]

the bond in a molecule of Cl2 is a covalent bond

6 0
2 years ago
4
Stolb23 [73]

Answer:

Um english please I don’t know Jewish

Explanation:

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3 years ago
1 mol super cooled liquid water transformed to solid ice at -10 oC under 1 atm pressure.
Arada [10]

Answer:

Explanation:

Given that:

number of moles of super cooled liquid water = 1

Melting enthalpy of ice = 6020 J/mol

Freezing point =0 °C = (0 + 273 K)= 273 K

The decrease in entropy of the system during freezing for 1 mol (i.e during transformation from liquid water to solid ice )  = - 6020 J/mol × 1 mol /273 K = -22.051 J/K

Entropy change during further cooling from 0 °C (273 K) to -10 °C (263 K)

\Delta \ S = \int\limits^{T_2}_{T_1}\dfrac{nC_p(s)dT}{T}

\Delta \ S = {nC_p(s)In \dfrac{T_2}{T_1}

\Delta \ S = {(1*37.7)In \dfrac{263}{273}

Δ S = -1.4 J/K

Total entropy change of the system = - 22.05 J/K - 1.4 J/K = - 23.45 J/K

Entropy change of universe = entropy change of the system+ entropy change of the surrounding

According to the second law of thermodynamics

Entropy change of universe  >0

SO,

Entropy change of the system + entropy change in the surrounding > 0

Entropy change in the surrounding > - entropy change of the system

Entropy change in the surrounding > - (- 23.53 J/K)

Entropy change in the surrounding > 23.53 J/K

b) Make some comments on entropy changes from the obtained data.

From the data obtained; we will realize that the entropy of the system decreases as cooling takes place when water is be convert to ice , randomness of these molecules reduces and as cooling proceeds , hence, entropy reduces more as well and the liberated heat will go into the surrounding due to this entropy of the surrounding increasing.

4 0
2 years ago
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