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.
The balanced chemical equation is given as:
2CH3CH2OH(l) → CH3CH2OCH2CH3(l) + H2O(l)
We are given the yield of CH3CH2OCH2CH3 and the amount of ethanol to be used for the reaction. These values will be the starting point for the calculations.
Theoretical amount of product produced:
329 g CH3CH2OH ( 1 mol / 46.07 g ) ( 1 mol CH3CH2OCH2CH3 / 2 mol CH3CH2OH ) (74.12 g / mol ) = 264.66 g CH3CH2OCH2CH3
% yield = .775 = actual yield / 264.66
actual yield = 205.11 g CH3CH2OCH2CH3
The grams of NaCl that are required to make 150.0 ml of a 5.000 M solution is 43.875 g
calculation
Step 1:calculate the number of moles
moles = molarity x volume in L
volume = 150 ml / 1000 = 0.15 L
= 0.15 L x 5.000 M = 0.75 moles
Step 2: calculate mass
mass = moles x molar mass
molar mass of NaCl = 23 + 35.5 = 58.5 mol /L
mass is therefore =0.75 moles x 58.5 mol /l =43.875 g
The end product will depend upon
a) the amount of the reagent taken
b) the final treatment of the reaction
If we have just taken methylmagnesium iodide and p-hydroxyacetophenone, then we will get methane and hydroxyl group substituted with MgI in place of hydrogen
Figure 1
However if we have taken excess of methylmagnesium iodide which is Grignard's reagent followed by hydrolysis we will get different product
Figure 2
1.54×10 −10
one and fifty four-hundreths times ten to the power of negitiive 10
