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3 years ago

Variation definition

Chemistry

1 answer:

mote1985 [20]3 years ago

4 0

Answer:

Explanation:

Variation problems involve fairly simple relationships or formulas, involving one variable being equal to one term. Here follows the most common kinds of variation.

The constant of variation in a direct variation is the constant (unchanged) ratio of two variable quantities. In the following equation y varies directly with x, and k is called the constant of variation:

y=kx

Another form of variation is the inverse variation which works when there is a relationship between two variables in which the product is a constant. When one variable increases the other decreases in proportion so that the product is unchanged.

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What did organizing information on the periodic table reveal?

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Answer:

The groupings of elements in the periodic table reveal various trends, present in groups such as the alkali metals and noble gases, or the lanthanides and actinides. The periodic table can be used to find information on elements or predict the properties of as-of-yet undiscovered elements.

Explanation:

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Do all forms of radioactive decay change the identity of the original element?

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2 years ago

Which describes a chemical reaction? A. Raw eggs are fried in a pan B. Raw eggs are scrambled with a fork C. Raw eggs are placed

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3 years ago

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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.

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3 years ago

An ionic compound has the general form, xcly. which option for x and y correctly completes the formula?

igor_vitrenko [27]

Answer:
Case 1:
  X =  Any element from Group I

i)   H
ii)   Li
iii) Na
   iv) K
v)   Rb
vi)  Cs

  Y =  1

Case 2:
  X = Any element from Group II

i) Be
ii) Mg
iii) Ca
iv) Sr
v) Ba
vi) Ra

  Y = 2

Case 3:
  X =  Any element from Group III

i)   B
ii)   Al
iii) Ga
   iv) In
v)   Ti

  Y =  3

Explanation:
The general formula given is as follow,

XCly

So, if X has +1 oxidation state, then it will require only one Cl atom with oxidation number -1 to form a neutral compound, therefore, y = 1.

If X has +2 oxidation state, then it will require two Cl atoms with oxidation number -1 to form a neutral compound, therefore, y = 2.

If X has +3 oxidation state, then it will require three Cl atoms with oxidation number -1 to form a neutral compound, therefore, y = 3.

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4 years ago