If a metal is bonded to carbon, they form ionic bonds. If a nonmetal forms a bond with carbon, they form covalent bond. Ionic bond requires metals to give away electrons to form cations and nonmetal to gain/accept electron to form anions. Covalent bond requires sharing of electrons between both elements. So if you meant exchange as in covalent bond, then no. Not many nonmetal elements form bonds with carbon as there are more metal elements in the periodic table.
Answer:
Because only regular-shaped objects have a consistent length, width, height, and shape.
Explanation:
Because only regular-shaped objects have a consistent length, width, height, and shape. When dealing with an object composed of differing shapes, your must calculate each portion separately and then add them.
Answer:
1. decomposition
2. combustion
3.single replacement
4. combination
5. double replacement
Explanation:
1. one compound is split into 2 elements
2. co2 and h20 was the product of the reaction
3. cu is replaced with co
4. 2 compounds become one compound
5. ca is replaced with na and na is replaced with ca
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%
