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Anvisha [2.4K]
4 years ago
14

Complete the passage.

Chemistry
2 answers:
mr Goodwill [35]4 years ago
8 0

Answer:

Complete the passage.

K (potassium) belongs to group IA of the periodic table, and has 1 valence electron. Br (bromine) belongs to VIIA, and has  _____ valence electrons.

The answer is <u>7</u>!!!!!

Crank4 years ago
4 0
Bromine has 7 valence electrons
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How does the mass of an oxygen nucleus
nataly862011 [7]

Answer: The mass of the nucleus is less than the mass of the 8 protons and 8 neutrons.

Explanation:

5 0
3 years ago
Of the two type of bonds (pi and sigma), which is capable of rotation around that bond and which is not? Why?
Bess [88]

Answer:

The sigma bonds are capable of rotating, the pi bonds not.

Explanation:

Sigma bonds are the strongest type of bonds, there are related to the overlapping in the atomic orbitals, and they can rotate. In the pi bound (that is a double bond), there are electrons moving on the molecule and it is not permitted the rotation on this type of bonds. Of the sigma bonds are capable of rotating while the pi bonds not.

8 0
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
3 years ago
For the endothermic reaction CaCO3 (s) CaO (s) CO2 (g) Le Cha^ telier's principle predicts that __________ will result in an inc
Firdavs [7]

Answer:

Increasing the temperature will result in increase in number of moles of CO_{2}

Explanation:

Concentration of pure solids and liquids remain constant throughout a reaction. Hence their concentrations terms are not involved in expression for equilibrium constant. Hence removing or adding solid CaCO_{3} will not affect change in number moles of CO_{2}.

In the forward direction, pressure increases due to formation of gaseous CO_{2}. Hence, according to Le-chatelier's principle, increase in pressure will result formation of more reactant.

As it is an endothermic reaction therefore heat is consumed in formation of CO_{2}. So, according to Le-chatelier;s principle, increasing the temperature will result in increase in number of moles of CO_{2}.

Correct option (D)

3 0
3 years ago
Read 2 more answers
Describe the key characteristics of a mixture using melting point, substances, composition, range, physical process, separated
Naily [24]
Well first you need to know what a mixure is. A mixture must be made of two or more elements or compounds. There are several ways to seperate a mixture, that is, to isolate each substance (element or compound). You can do this with melting point (i.e. a block of aluminum and iron that have been melted, mixed and then cooled. To seperate the Al from the Fe, you can start to melt it. The Al melts at a much lower temp than the Fe, and therefore will liquify before Fe.) 
3 0
3 years ago
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