Answer:
2OH^-(aq) + Cu^2+(aq) -----> Cu(OH)2(s)
Explanation:
The net ionic equation usually shows the main ionic reaction that goes in the system. The other ions that do not participate in this net ionic equation are called spectator ions. Spectator ions do not participate in the main reaction occurring in the system.
The net ionic equation quite often result in the formation of a solid precipitate in the system such as Cu(OH)2.
The net ionic equation for this reaction is;
2OH^-(aq) + Cu^2+(aq) -----> Cu(OH)2(s)
Answer: Hello the compound is missing but I was able to get the Full question and missing compound . ( compound = copper sulfate )
<em>answer</em> : statement ; 2 , 3 and 5
Explanation:
The true statements regarding the coordination compound ( copper sulfate ) are :
- The ligand must have at least one unshared pair of valence electrons in order to covalently bond with transition metal in the coordination compound ( statement 2 )
- Ethanol was used during crystallization of the coordination compound because the compound is soluble in ethanol ( statement 3 )
- The colors of many coordination compounds are the result of light absorption by the d electrons on the transition metal ( statement 5 )
During the coordination of compounds dative bonds exits between the transition metals and the Ligands molecules
I think the best answer is the last option. A scatter plot is the appropriate type of graph for the student to use to show the percent samples per group. This plot is somewhat similar to line graphs. However, they are use for a specific purpose which is to show the relationship between two parameters. In this case, the correlation between pH and the percent of samples.
<span> </span><span>1. Alcohol(other)
3. Acid
5. Salt
</span>
Explanation:tr
a) Molar mass of HF = 20 g/mol
Atomic mass of hydrogen = 1 g/mol
Atomic mass of fluorine = 19 g/mol
Percentage of an element in a compound:
Percentage of fluorine:
Percentage of hydrogen:
b) Mass of hydrogen in 50 grams of HF sample.
Moles of HF = 
1 mole of HF has 1 mole of hydrogen atom.
Then 2.5 moles of HF will have:
of hydrogen atom.
Mass of 2.5 moles of hydrogen atom:
1 g/mol × 2.5 mol = 2.5 g
2.5 grams of hydrogen would be present in a 50 g sample of this compound.
c) As we solved in part (a) that HF molecules has 5% of hydrogen by mass.
Then mass of hydrogen in 50 grams of HF compound we will have :
5% of 50 grams of HF = 
