"<span>using shaving cream, rather than shaving on dry skin" I believe this is the answer. coz rest of them would increase friction rather than decrease</span>
Answer:
The correct matching of the air mass and the letters in the word bank are given as follows;
1. Warm and humid ↔ D
2. Extremely cold and dry ↔ B
3. Cold and dry ↔ A
4. Cold and humid ↔ C
5. Warm and dry ↔ E
Where;
A Represents continental polar
B Represents Artic
C Represents Maritime Polar
D Represents Maritime Tropical
E Represents Continental Tropical
Explanation:
A. A continental polar is one that can be described as a cold and dry climate as the region is located at a further away from the oceanic water bodies that add humidity to the climate
B. The regions of the Artic and the Antarctic have very limited amount of precipitation every year because the air is very cold as well as dry
C. A polar climate is a cold climate region, while a maritime climate is humid.
Therefore, the maritime polar climate combines both cold and humid conditions
D. A warm and humid region has high rainfall and humidity, as such the maritime climate which are humid and the tropical climate, which are warm, combine to give a warm and humid climate
E. The continental tropical climate can be described as warm and dry, compared to the continental water bodies, due to the location being distant from and therefore, the absence of high moisture containing wind that comes from the oceans.
<u>Answer:</u> The value of 
for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K
<u>Explanation:</u>
Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.
The equation used to calculate entropy change is of a reaction is:
![\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20S%5Eo_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20S%5Eo_%7B%28reactant%29%7D%5D)
For the given chemical reaction:
The equation for the entropy change of the above reaction is:
![\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20S%5Eo_%7B%28NO_2%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20S%5Eo_%7B%28N_2%28g%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20S%5Eo_%7B%28O_2%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta S^o_{rxn}=[(2\times (240.06))]-[(1\times (191.61))+(2\times (205.14))]\\\\\Delta S^o_{rxn}=-121.77J/K](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%28240.06%29%29%5D-%5B%281%5Ctimes%20%28191.61%29%29%2B%282%5Ctimes%20%28205.14%29%29%5D%5C%5C%5C%5C%5CDelta%20S%5Eo_%7Brxn%7D%3D-121.77J%2FK)
Entropy change of the surrounding = - (Entropy change of the system) = -(-121.77) J/K = 121.77 J/K
We are given:
Moles of nitrogen gas reacted = 1.90 moles
By Stoichiometry of the reaction:
When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K
So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = 
Hence, the value of for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K
It will be number 1.
NO loses oxygen ie its being reduced to N2
NH3 gains oxygen atom ie its being oxidised to H2O
Answer:
A molecule is a group of two or more atoms held together by chemical bonds. A compound is a substance which is formed by two or more different types of elements which are united chemically in a fixed proportion. All molecules are not compounds.
Good luck !
