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riadik2000 [5.3K]
3 years ago
6

Name the compound formed when a copper cation has a 2+ charge combined with a bromine anion that has a 1- charge.

Chemistry
1 answer:
Rashid [163]3 years ago
7 0

Answer: The compound is called copper (II) bromide or cupric bromide

Explanation:

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Answer:

A.

Explanation:

A redox reaction is a reaction when oxidation states (or numbers) change during reaction.

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3 years ago
Draw it draw lewis dot structures for each hypothetical molecule shown below, using the correct number of valence electrons for
soldi70 [24.7K]

The first molecule is a sensible molecule having complete octet of each atom such as C, H and O whereas the second molecule having hydrogen present between the aldehyde and methyl group and thus showing hydrogen is making bond with aldehyde and methyl as well which is not possible because hydrogen only having one electron in its octet due to which it can only form a single bond by sharing its valence electron.

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3 years ago
I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r
maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P

Now, the moles of Li3P consumed by 15 g of Al2O3:

n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP

Therefore, the total mass of products is:

m_{products}=13g+8.5g\\\\m_{products}=21.5g

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

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3 years ago
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