We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.
First, we need to determine the required moles of CaCl₂. We have 500 mL (0.500 L) of a 0.360 M solution (0.360 moles of CaCl₂ per liter of solution).
Then, we will convert 0.180 moles to grams using the molar mass of CaCl₂ (110.98 g/mol).
To prepare the solution, we weigh 20.0 g of CaCl₂ and add it to a beaker with enough distilled water to dissolve it. We stir it, heat it if necessary, and when we have a solution, we transfer it to a 500 mL flask and complete it to the mark with distilled water.
We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.
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Answer: The product from the reduction reaction is
CH3-CH2-CH(CH3)-CH2-CH2OH
IUPAC name; 3- Methylpentan-1-ol
Explanation:
Since oxidation is simply the addition of oxygen to a compound and reduction is likewise the addition of hydrogen to a compound.
Therefore, hydrogen is added onto the carbon atom adjacent to oxygen in 3- methyl pentanal
CH3 CH2 CHCH3 CH2 CHO thereby -CHO( aldehyde functional group) are reduced to CH2OH ( Primary alcohol) which gives;
3-methylpenta-1-ol .
The structure of the product is:
CH3-CH2-CH(CH3)-CH2-CH2OH
Answer:
If You Breathe Fast, CO2 Is Blank The Equilibrium Shifts To Blank [H3O+], Which Raises The PH.
Explanation:
Answer:
The main purpose of the expedition was to conduct a hydrographic survey of the coasts of the southern part of South America.
Boyle's Law states: pV = constant.
24.43 x 1.895 = 46.29485
therefore, 15.6 x _____ = 46.29485
unknown = 2.968L
