The term formula units means molecules.
Then, what you are looking for is the mass in 4.59*10^24 molecules.
The procedure involves to convert the 4.59 * 10^24 molecules into moles and use the molar mass of the sodium chloride.
1) Number of moles = 4.59 * 10^24 molecules / (6.02 * 10^23 molecules/mol) = 7.62 mol
2) Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol
3) mass of NaCl = molar mass * number of moles = 58.44 g/mol * 7.62 mol = 445.31 g of NaCl
Answer: 445.31 g of NaCl.
The name of the compound by using the <u>IUPAC nomenclature of organic compounds</u> is 1 -octene. The correct option is the last option - 1-octene.
<h3>Nomenclature of Organic compounds</h3>
From the question, we are to determine the name of the given molecule.
To name the compound, we will follow the IUPAC rules.
Some of IUPAC rules are
- Find the longest continuous carbon chain. Determine the root name for this parent chain.
- For Alkenes (organic compounds with double bond), number the chain of carbons that includes the C=C so that the C=C has the lower position number. Change “ane” to “ene” and assign a position number to the first carbon of the C=C.
The given compound has 8 carbons and a double bond. The root name of the compound is octane.
By <u>IUPAC rules</u>, the compound is an <u>Octene</u>.
Since the double bond is between carbon-1 and carbon-2. The compound becomes 1-octene.
Hence, the name of the compound by using the <u>IUPAC nomenclature of organic compounds</u> is 1 -octene. The correct option is the last option - 1-octene.
Learn more on Nomenclature of Organic compounds here: brainly.com/question/26754333
The diagram for the compound is attached below.
Answer:
HCl conc.= 6.0mol/L
Explanation:
From the dissociation of HCl= 1 mole H+ and 1mol Cl-, which is equivalent stoichiometrically in concentration to that of 1 mol HCl,
Option 1/A (It is the first one)
Explanation:
We will balance equation which describes the reaction between sulfuric acid and sodium bicarbonate: as follows.
Next we will calculate how many moles of
are present in 85.00 mL of 1.500 M sulfuric acid.
As, Molarity = 
1.500 M = 
n = 0.1275 mol
Now set up and solve a stoichiometric conversion from moles of
to grams of
. As, the molar mass of
is 84.01 g/mol.
= 21.42 g
So unfortunately, 15.00 grams of sodium bicarbonate will "not" be sufficient to completely neutralize the acid. You would need an additional 6.42 grams to complete the task.