Answer:
The correct answer is because the molecular structure.
Explanation:
The difficulty of ammonia and methane to be represented on paper is due to the molecular structure. These compounds have a three-dimensional projection with defined angles. Ammonia presents angles of 109.5º between the atom of Nitrogen and those of Oxygen. The ammonia presents 107.8º between the oxygen atoms.
In the methane molecule, there is 109.5º between the hydrogen molecules and the carbon atom. This results in the need for a 3D representation of the molecule.
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Answer:
Try this link https://chem.libretexts.org/Courses/Mount_Aloysius_College/CHEM_100%3A_General_Chemistry_(O'Connor)/08%3A_Solids_Liquids_and_Gases/8.E%3A_Solids_Liquids_and_Gases_(Exercises)
Answer:
"Heat of combustion"
Explanation:
A physical property is characteristic of a substance that can be observed or measured without changing its chemical composition, such as color, molecular weight, and volume. A chemical property can only be observed or measured by changing the chemical composition of the substance, such as reactivity, flammability and oxidation states.
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Answer:
The percent by mass of copper in the mixture was 32%
Explanation:
The ammount of HNO₃ used is:
mol HNO₃ = volume * concentration
mol HNO₃ = 0.015 l * 15.8 mol/l
mol HNO₃ = 0.237 mol
According to the reaction, 4 mol HNO₃ will react with 1 mol Cu and produce 1 mol Cu²⁺. Since we have 0.237 mol HNO₃, the amount of Cu that could react would be (0.237 mol HNO₃ * 1 mol Cu / 4 mol HNO₃) 0.06 mol. This reaction would produce 0.060 mol Cu²⁺, however, only 0.010 mol Cu²⁺ were obtained, indicating that only 0.010 mol Cu were present in the mixture. This means that the acid was in excess, so we can assume that all copper present in the mixture has reacted.
Since 0.010 mol of Cu²⁺ were produced, the amount of Cu was 0.01 mol.
1 mol of Cu has a mass of 63.5 g, then 0.01 mol has a mass of:
0.01 mol Cu * 63.5 g / 1 mol = 0.635 g.
Since this amount was present in 2.00 g mixture, the amount of copper in 100 g of the mixture will be:
100 g(mixture) * 0.635 g Cu / 2 g(mixture) = 32 g
Then, the percent by mass of Cu (which is the mass of Cu in 100 g mixture) is 32%
Explanation:
In general, to balance an equation, here are the things we need to do:
Count the atoms of each element in the reactants and the products.
Use coefficients; place them in front of the compounds as needed.