The percent by mass of carbon in acetic acid (HC2H3O2) is calculated using the molar masses of its constituent elements. Upon calculation, it is determined that the percent by mass of carbon in acetic acid is 40.00%.
To calculate the percent by mass of carbon in acetic acid (HC2H3O2), we need to use the molar masses of the elements involved. The formula for acetic acid includes two carbon atoms, four hydrogen atoms, and two oxygen atoms. Using the given molar masses (carbon: 12.01 g/mol, hydrogen: 1.008 g/mol, oxygen: 16.00 g/mol), the molar mass of acetic acid can be calculated as follows:
Adding these together gives the molar mass of acetic acid as 60.052 g/mol.
To find the percent by mass of carbon, we take the total mass of carbon in the compound and divide it by the total molar mass of the compound:
(24.02 g/mol / 60.052 g/mol) × 100% = 40.00%
Therefore, the percent by mass of carbon in acetic acid is 40.00%.
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Answer:
Conversion gives 42,107 km in total
Explanation:
A marathon covers 26 miles and 285 yards.
If 1 kilometer is 0.6214 miles, then how many kilometers are 26 miles, then:
1 km = 0.6214 miles
X = 26 miles
X = 41,841 km
Now if 1 yard is 36 inches, how many inches is 285 yards?
1 yard = 36 inches
285 yards = X
X = 10260 inches
If an inch is equal to 2.54 cm, how many centimeters are 10260 inches?
1 inch = 2.54 cm
10260 cm = X
X = 26060.4 cm
And 26060.4 cm is 0.260604 km
Finally, the sum of the distance would be:
41.841 km + 0.260604 km = 42.107 km
FALSE
Answer:
True
Explanation:
Amorphous solid is the designation given to a structure that has no long-range (atomic) spatial ordering, such as regular solids. For a material to be considered as regular solid it must have a defined crystalline structure, which corresponds to a positioning of the atoms that compose it. Amorphous materials have no definite geometric shape, may even have some rigidity like ordinary solids, but their atoms are not in spatial organization (at the microscopic level), in which case amorphous solids have no general order, with individual clusters joined together only. by the forces of Van der Waals.
Answer:
342.17 g/mol
Explanation:
Molecular formulas is the actual number of atoms of each element in the compound. Molecular mass is the sum of the mass of each element which is present in the molecular formula is multiplied by their subscript which appear in the molecular formula.
Given,
The formula is =
Mass from the formula = 2×Molar mass of Aluminium + 3×Molar mass of sulfur + 12×Molar mass of Oxygen
Molar mass of Aluminium = 27.00 g/mol
Molar mass of sulfur = 32.06 g/mol
Molar mass of Oxygen = 15.99 g/mol
So,
Mass from the formula = 2×27.00 + 3×32.06 + 12×15.99 g/mol ≅ 342.17 g/mol
The molar mass of is 342.17 g/mol. The correct answer is option C.
Molar mass is the mass of one mole of a substance. It is expressed in units of grams per mole (g/mol).
To calculate the molar mass of , we need to add up the atomic masses of all the atoms in one formula unit.
So, the molar mass of can be calculated as follows:
Molar mass = (2 × atomic mass of Al) + (3 × atomic mass of S) + (12 × atomic mass of O)
= (2 × 26.98 g/mol) + (3 × 32.07 g/mol) + (12 × 16.00 g/mol)
= 342.17 g/mol
Therefore, the answer is option C. 342.17 g/mol is the molar mass of .
Learn more about molar mass here:
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the known density of the liquid is
the density of a liquid
Thus, experimental error of 0.0108 g /mL is present.
Thus, Percentage error of 7.31 % is present.
b. The density is the same throughout the entire mixture.
c. It is not possible to separate the various parts of the mixture.
d. Taking one sample will enable you to determine what it is made of.
Answer: Option (a) is the correct answer.
Explanation:
A heterogeneous mixture is defined as the mixture in which distribution of solvent particles is uneven throughout the solution.
For example, a mixture of sand and water is a heterogeneous mixture.
So, when we take different samples for the same heterogeneous mixture then there will be different composition of solute particles in each of them.
On the other hand, in a homogeneous solution there will be even distribution of solute particles into the solvent. Each sample of a homogeneous mixture will contain same composition of solute particles.
Thus, we can conclude that the statement different samples may not be exactly alike, is true of a heterogeneous mixture.
(1) 6 (2) 8 (3) 16 (4) 32
Answer: it’s 16
Explanation: trust me on this one