First, you need to count copper mass in alloy.
Second, you have to make an equation an find x ( the copper mass must be added). The answer is: 13,5g pure copper
Answer:
9.62moles of NH₃
Explanation:
Given parameters:
Number of moles of nitrogen = 4.81moles
Unknown:
Number of moles of NH₃ = ?
Solution:
To solve this problem, we need to establish a balanced reaction equation:
N₂ + 3H₂ → 2NH₃
Now, we can solve the problem by working from the known to the unknown.
1 mole of N₂ produced 2 moles of NH₃
4.81moles of N₂ will produce 2 x 4.81 = 9.62moles of NH₃
Answer:
1) the oxidation state of the metal is +2
2) there are four d electrons
3) metal valence electrons =4 ligand valence electrons= 18
4) sp^3d^2
5) octahedral geometry
Explanation:
[Cr(H2O)6]2+ is an octahedral complex. Octahedral complexes are known to have the metal ion in sp^3d^2 hybridization state. Since there are six ligands each bonding to the central metal ion, there are eighteen ligand valence electrons and four valence electrons from the metal present in the high spin complex.
The crystal field of high spin and low spin octahedral for a d4 ion is shown in the image attached.
Answer:
the two spectator ions are; K(+) and NO3(-)
Explanation:
First off, let's write out the balanced chemical equation for the reaction;
3K2CO3(aq) +2Fe(NO3)3(aq) ----> 6KNO3(aq) + Fe2(CO3)3(s)
In order to identify which ions are spectators, we have to break the equation down to an ionic equation. This is done by splitting all aqueous compounds into ions while leaving the solids, liquids as they are.
We have;
K(+) + CO3(2-) + Fe(3+) + NO3(-) ---> K(+) + NO3(-) + Fe2(CO3)3(s)
Spectators ions are pretty much those ions that do not undergo a change in the reaction. Spectator ions always have the same number of moles and charge in both sides of the reaction.
Upon observing the ionic equation, we can tell that the two spectator ions are; K(+) and NO3(-)
It can be shown that matter is neither created nor destroyed in a chemical reaction by taking a critical look at the each atom in both sides of the reaction (reactant and product sides).
1 atom of Mg started the reaction and 1 atom is also present in the products.
8 atoms of O are present in the reactants and 8 is also present in the products.
4 H atoms are present in the reactants and 4 are also present in the products.
2 N atoms are present in the reactants and 2 are also in the products.
In other words, the total number of each atom that is present in the reactants are also present in the product. Nothing has been lost during the reaction, although, the forms of each atom might have changed.
More on the law of conservation of matter can be found here: brainly.com/question/20635180
