The mass of trifluoroacetic acid that contains 83.3 g of oxygen is approximately 482.6 g.
How to calculate the mass of trifluoroacetic acid (C2HF3O2) containing 83.3 g of oxygenFirst we need to use the molar mass and stoichiometry of the compound.
The molar mass of trifluoroacetic acid is:
(2 x 12.01 g/mol) + (1 x 19.00 g/mol) + (3 x 19.00 g/mol) + (2 x 16.00 g/mol) = 114.02 g/mol
The molar mass of oxygen is 16.00 g/mol.
From the chemical formula of trifluoroacetic acid (C2HF3O2),
we can see that each mole of the compound contains 2 moles of carbon, 1 mole of hydrogen, 3 moles of fluorine, and 2 moles of oxygen.
Using this information, we can set up a proportion to find the mass of trifluoroacetic acid that contains 83.3 g of oxygen:
(83.3 g O2) x (1 mol O2/16.00 g O2) x (1 mol C2HF3O2/2 mol O2) x (114.02 g C2HF3O2/1 mol C2HF3O2) = 482.6 g
Therefore, the mass of trifluoroacetic acid that contains 83.3 g of oxygen is approximately 482.6 g.
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Which characteristic is not a periodic trend?
A. Atomic radius
B. Electron affinity
C. Ionization energies
D. Number of elements in a group
Answer:
Explanation:
D
Add electrons and the charge for the resulting radical
cationic intermediate, and add curved arrows for the
a-cleavage mechanism en route to m/z 43.
Electrons are negatively charged subatomic particles that orbit the nucleus of an atom. They are essential for chemical reactions, as they participate in the formation and breaking of chemical bonds. The appropriate diagram relating to the question is attached.
How to convey the informationA radical is a molecule or atom that contains one or more unpaired electrons. Radicals are highly reactive and can participate in a wide range of chemical reactions, such as combustion, polymerization, and oxidation.
A cationic intermediate is a positively charged species that forms during a chemical reaction. It is an intermediate state between the starting materials and the final products, and it is typically short-lived. Cationic intermediates can be formed by the loss of an electron or by the addition of a proton or other positively charged species.
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Determine the empirical formula of a compound containing 47. 37 grams of carbon, 10. 59 grams of hydrogen, and 42. 04 grams of oxygen.
the empirical formula of a compound containing 47. 37 grams of carbon, 10. 59 grams of hydrogen, and 42. 04 grams of oxygen is C3H8O2.
To determine the empirical formula of the compound, we need to find the ratios of the different elements present in the compound.
The first step is to convert the masses of each element to moles using their respective molar masses:
Carbon: 47.37 g / 12.01 g/mol = 3.94 mol
Hydrogen: 10.59 g / 1.01 g/mol = 10.48 mol
Oxygen: 42.04 g / 16.00 g/mol = 2.63 mol
Next, we divide each of the mole values by the smallest of the three, which is 2.63 mol:
Carbon: 3.94 mol / 2.63 mol = 1.50
Hydrogen: 10.48 mol / 2.63 mol = 3.98
Oxygen: 2.63 mol / 2.63 mol = 1.00
Now we need to convert these ratios to whole numbers by multiplying each by a common factor. The smallest ratio is 1.00, so we will multiply all the ratios by 2 to get:
Carbon: 1.50 x 2 = 3
Hydrogen: 3.98 x 2 = 8
Oxygen: 1.00 x 2 = 2
Therefore, the empirical formula of the compound is C3H8O2.
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Given the following equation: 2N₂+502 → 2N₂O5
How many moles of N₂O5 are produced by reacting 10 moles of O₂?
We can form mole ratios by using a balanced chemical equation.
Solving the Question[tex]2N_2+5O_2\rightarrow 2N_2O_5[/tex]
2 moles of [tex]N_2O_5[/tex] can be formed with 5 moles of [tex]O_2[/tex]:[tex]\dfrac{n_{N_2O_5}}{2}=\dfrac{n_{O_2}}{5}[/tex]
We're given that we have 10 moles of [tex]O_2[/tex]:[tex]\dfrac{n_{N_2O_5}}{2}=\dfrac{10}{5}[/tex]
[tex]n_{N_2O_5}=4[/tex]
Answer4 moles of N2O4 are produced by reacting 10 moles of O2.
As per the balanced equation of the reaction, 5 moles of oxygen molecule gives 2 moles of the product N₂O₅ . Hence, 10 moles of oxygen will give 4 moles of nitrogen pentoxide.
What is mole ratio ?The balanced chemical equation of a reaction represents the perfect stoichiometry of the reactants and products. Hence, the number of atoms of each element in the reactant side must be equal to their number of atoms in the product side.
The stochiometric ratio of number of moles of one reactant to the other or the product is called its mole ratio.
From the given reaction, it is clear that 5 moles of oxygen molecules reacts with nitrogen gas to give 2 moles of dinitrogen pentoxide.
Hence, the number of moles of dinitrogen pentoxide produced by 10 moles of oxygen gas is calculated as follows:
(10× 2)/5 = 4
Therefore, the number of moles of the product formed from 10 moles of oxygen gas is 4 moles.
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