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Alisiya [41]
3 years ago
9

Why do we observe national day​

Chemistry
1 answer:
WITCHER [35]3 years ago
5 0

A national day is a day on which celebrations mark the nationhood of a nation or state. It may be the date of independence, of becoming a republic, or a significant date for a patron saint or a ruler (such as a birthday, accession, or removal). The national day is often a public holiday.

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How many carbon atoms are in 7.05 moles of pyridine
NemiM [27]
First, it is best to know the chemical formula of pyridine which is C5H5N. To determine the number of carbon atoms present in pyridine, multiply 7.05 mol C5H5N with 5 mol C/ 1 mol C5H5N which then results to 35.35 mol of carbon. Then, multiply the answer to Avogadro's number which is 6.022x10^23 atoms. It is then calculated that the number of carbon atoms in 7.05 moles of pyridine is 2.12x10^25 atoms. 
5 0
3 years ago
Read 2 more answers
During photosynthesis, ________.
ludmilkaskok [199]

Answer:

d. there is a net consumption of water and carbon dioxide

Explanation:

Photosynthesis, is the process whereby light energy is transform into chemical energy by

green plants and other photosynthesis capable organisms . In the process of photosynthesis, light energy is captured by green plants which it uses to convert carbon dioxide water, and minerals into energy-rich organic compounds and oxygen is evolved as a byproduct.

It is a chemical reaction taking place inside a plant, resulting in the production of food for the survival of the plant.

Photosynthesis takes place in the leaves of a plant in the presence of sunlight and.

5 0
3 years ago
What is the molarity of a 0.65L solution containing 63 grams of ? The molar mass of NaCl is 58.44 g/mol
ad-work [718]

Explanation:

The molarity of a solution is defined like the number of moles of solute per liters of solution.

molarity = moles of solute/(volume of solution in L)

We know the volume of solution in L.

volume of solution = 0.65 L

To go from the mass of our solute in grams to moles we have to use its molar mass.

mass of NaCl = 63 g

molar mass of NaCl = 58.44 g/mol

moles of NaCl = 63 g * 1 mol/(58.44 g)

moles of NaCl = 1.078 moles

Finally we can find the molarity of the solution

molarity = moles of NaCl/(volume of solution)

molarity = 1.078 moles/(0.65 L)

molarity = 1.66 M

Answer: the molarity of the solution is 1.66 M.

7 0
1 year ago
Select all statements that correctly describe the reactions of benzene. A. Benzene typically undergoes reactions in which the ar
Katarina [22]

Answer:Benzene typically undergoes reactions in which the aromatic ring is preserved.B. Benzene typically reacts with electrophiles where an aromatic proton is substituted by the electrophile

Explanation:

The reactions of benzene are such that the aromatic ring is not destroyed. Addition reactions destroy the aromatic ring hence they aren't typical reactions of benzene. Benzene rings are attacked by electrophiles in which reaction a proton is substituted by the electrophile. Alkenes only undergo addition reaction and not electrophilic substitution reaction.

8 0
3 years ago
The formation of tert-butanol is described by the following chemical equation: (CH3), CBr (aq) + OH(aq) → Br" (aq) +(CH), COH(aq
Dafna11 [192]

Answer:

(CH_3)_3С^+ (aq) + OH^- (aq)\rightarrow (CH_3)_3COH (aq)

Explanation:

There are two ways of looking at this problem. The first way, slightly more advanced, is to understand that the carbocation formed is an intermediate in this reaction: it is formed in one step and consumed in the subsequent step.

Secondly, we have hydroxide involved as our reactant, so it should be our second reactant in the second bimolecular step.

Thirdly, the product formed would be a combination of the anion and cation, one of our products, this means we have the following second step:

(CH_3)_3С^+ (aq) + OH^- (aq)\rightarrow (CH_3)_3COH (aq)

Another way is to verify this knowing that by adding all of the steps should yield a net equation, notice if we add the two steps together (reactants on one side and products on the other), we obtain:

(CH_3)_3С^+ (aq) + OH^- (aq) + (CH_3)_3CBr (aq)\rightarrow (CH_3)_3COH (aq) + (CH_3)_3C^+ (aq) + Br^- (aq)

Notice that the intermediate carbocation cancels out on both sides to yield the final net equation:

OH^- (aq) + (CH_3)_3CBr (aq)\rightarrow (CH_3)_3COH (aq) + Br^- (aq)

This means we have the correct second step.

6 0
3 years ago
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