Answer:
Concentration of chloride ions = 0.584M
Explanation:
The step by step calculations is shown as attached below.
Answer:
There are five signs of a chemical change:
Colour Change.
Production of an odour.
Change of Temperature.
Evolution of a gas (formation of bubbles)
Precipitate (formation of a solid).
Answer:
The molarity of urea in this solution is 6.39 M.
Explanation:
Molarity (M) is <em>the number of moles of solute in 1 L of solution</em>; that is

To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.
Our first step is to calculate the moles of urea in 100 grams of the solution,
using the molar mass a conversion factor. The total moles of 100g of a 37.2 percent by mass solution is
60.06 g/mol ÷ 37.2 g = 0.619 mol
Now we need to calculate the volume of 100 grams of solution, and we use density as a conversion factor.
1.032 g/mL ÷ 100 g = 96.9 mL
This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.
0.619 mol/96.9 mL × 1000 mL= 6.39 M
Therefore, the molarity of the solution is 6.39 M.
Within the options, we have four organic compounds. Let's see what the skeletal structure of the compounds is in order to identify them better:
The first compound CH3OCH3 has two methyl groups linked by a carbon atom, this type of compound is called an Ether
The second compound has a double bond, it is badly written but it seems that is an alkene.
The third compound has two methyl groups linked by nitrogen atoms, therefore will be an amine.
The last compound has a hydroxyl group, therefore it is an alcohol
Answer:
CH3OCH3 Ether
CH2CH2CHCH2CH3 Alkene
CH3NHCH3 Amine
CH3CH(OH)CH₂CH3 Alcohol
<span>Synthesis, decomposition, single replacement and double replacement.</span>