Ammonia is colorless gas with a characteristic smell. Its density is 0.589 times than air which makes it lighter than air. Ammonia can be easily liquefied due to the hydrogen bonding between the molecules. The boiling point is at -33.3 degrees Celsius and the freezing point is at -77.7 degrees Celsius.
Actually, we can answer the problem even without the first statement. All we have to do is write the reaction for the production of sulfur trioxide.
2 S + 3 O₂ → 2 SO₃
The stoichiometric calculations is as follows:
7 g S * 1 mol/32.06 g S = 0.218 mol S
Moles O₂ needed = 0.218 mol S * 3 mol O₂/2 mol S = 0.3275 mol O₂
Since the molar mas of O₂ is 32 g/mol,
Mass of O₂ needed = 0.3275 mol O₂ * 32 g/mol = 10.48 g O₂
The molality of the solution is 0.00037 m.
<h3>What is concentration?</h3>
The term concentration refers to the amount of solute in a solution.
We have the following information;
Molarity = 0.335 M
Density = 1.0432 g/mL
Temperature = 20 o C
The molality of the solution is obtained from;
m = 0.335 M × 1.0432 g/mL/ 1000(1.0432 g/mL) - 0.335 M (342 g/mol)
m = 0.344/1043.2 - 114.57
m = 0.344/928.63
m = 0.00037 m
Learn more about molality of solution: brainly.com/question/4580605
Answer:
At the end of meiosis, there are four cells, each with 23 chromosomes, for a total of 92 chromosomes split between the four cells.
Explanation:
During meiosis, a diploid cell (46 chromosomes) replicates its DNA (making 92 chromosomes) then undergoes two cell divisions to generate four haploid cells (23 chromosomes).
These haploid cells are the gametes which, during fertilization, fuse to become a zygote with 46 chromosomes.
Answer:
See explanation below
Explanation:
First, we need to understand that the monochlorination of an alkane like this one, involves substitution of one of the atoms of hydrogen of the molecule for an atom of chlorine.
This reaction takes place when the alkane reacts with Cl₂ in presence of light or heat.
When this happens, the first step involves the breaking of the double bond of the chlorine to form the ion Cl⁻.
The next step involves the substraction of the hydrogen of the molecule by the Chlorine. This will leave the alkane with a lone pair available for reaction.
The third step, the alkane with the lone pair of electron substract a chlorine for the beggining and form the mono chlorinated product.
The final step involves forming the remaining products with the remaining reagents there.
In the picture attached you have the mechanism and product for this reaction:
