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
H2SO4 is referred to as a strong acid and is denoted as option A.
<h3>What is an Acid?</h3>
This refers to any substance which tastes sour when in water and changes the color of blue litmus paper to red. It is usually very corrosive and are used in industries for different functions.
H2SO4 is referred to as a strong acid because it dissociates completely in its aqueous solution or water.
Read more about Acid here brainly.com/question/25148363
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Answer:
turgor pressure can be done in a lab or a self test.
turgor pressure is key to the plant’s vital processes. It makes the plant cell stiff and rigid. Without it, the plant cell becomes flaccid. Prolonged flaccidity could lead to the wilting of plants.
Turgor pressure is also important in stomate formation. The turgid guard cells create an opening for gas exchange. Carbon dioxide could enter and be used for photosynthesis. Other functions are apical growth, nastic movement, and seed dispersal.
Explanation:
- salt is bad for turgor pressure.
- Turgidity helps the plant to stay upright. If the cell loses turgor pressure, the cell becomes flaccid resulting in the wilting of the plant.
- The wilted plant on the left has lost its turgor as opposed to the plant on the right that has turgid cells.
The amount
per 100 g is:
38.7 %
calcium = 38.7g Ca / 100g compound = 38.7g
19.9 %
phosphorus = 19.9g P / 100g compound = 19.9g
41.2 %
oxygen = 41.2g O / 100g compound = 41.2g
The molar amounts of calcium,
phosphorus and oxygen in 100g sample are calculated by dividing each element’s
mass by its molar mass:
Ca = 38.7/40.078
= 0.96
P = 19.9/30.97
= 0.64
O = 41.2/15.99
= 2.57
C0efficients
for the tentative empirical formula are derived by dividing each molar amount
by the lesser value that is 0.64 and in this case, after that multiply wih 2.
Ca = 0.96 /
0.64 = 1.5=1.5 x 2 = 3
P = 0.64 /
0.64 = 1 = 1x2= 2
O = 2.57 /
0.64 = 4= 4x2= 8
Since, the
resulting ratio is calcium 3, phosphorus 2 and oxygen 8
<span>So, the
empirical formula of the compound is Ca</span>₃(PO₄)₂
