Answer:
57.6g
Explanation:
So, if in one mole of water, 16 g of oxygen atom is present. Then, in 3.6 moles of water, the mass of oxygen present will be 3.6×16=57.6g. Therefore, the amount of oxygen present in 3.6 g water is option (B)- 57.6 g.
Answer:
See explanation
Explanation:
The question is incomplete because the image of the alcohol is missing. However, I will try give you a general picture of the reaction known as hydroboration of alkenes.
This reaction occurs in two steps. In the first step, -BH2 and H add to the same face of the double bond (syn addition).
In the second step, alkaline hydrogen peroxide is added and the alcohol is formed.
Note that the BH2 and H adds to the two atoms of the double bond. The final product of the reaction appears as if water was added to the original alkene following an anti-Markovnikov mechanism.
Steric hindrance is known to play a major role in this reaction as good yield of the anti-Markovnikov like product is obtained with alkenes having one of the carbon atoms of the double bond significantly hindered.
The density of the liquid is 0.98 g/mL
<h3>What is density? </h3>
The density of a substance is defined as the mass of the subtance per unit volume of the substance. Mathematically, it can be expressed as:
Density = mass / volume
With the above formula, we can obtain the density of the liquid.
<h3>How to determine the density </h3>
- Mass = 30.8 g
- Volume = 31.5 mL
- Density =?
Density = mass / volume
Density of liquid = 30.8 / 31.5
Density of liquid = 0.98 g/mL
Learn more about density:
brainly.com/question/952755
Answer: Option B. 76.83L
Explanation:
1 mole of a gas occupy 22.4L at stp. This implies that 1mole of Radon also occupy 22.4L at stp.
If 1 mole of Radon = 22.4L
Therefore, 3.43 moles of Radon = 3.43 x 22.4 = 76.83L
