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Oxana [17]
4 years ago
8

A neutral lithium atom has 1 valence electron. A neutral bromine atom has 7 valence electrons. A chemical reaction between the 2

forms lithium bromide.
What electrical charges would the 2 newly formed ions take on?
A. Li- and Br-
B. Li+ and Br-
C. Li- and Br+
D. Li+ and Br+
Chemistry
2 answers:
Sholpan [36]4 years ago
4 0

Answer: Option (B) is the correct answer.

Explanation:

Atomic number of lithium is 3 and its electronic distribution is 2, 1.

Atomic number of bromine is 35 and its electronic distribution is 2, 8, 18, 7.

It is known that a neutral atom which accepts electrons acquires a negative charge and a neutral atom that loses electrons will acquire a positive charge.

So, in order to attain stability lithium will lose its 1 valence electron and bromine will accept its 1 electron.

Therefore, lithium will acquire a positive charge and bromine will acquire a negative charge.

Thus, we can conclude that electrical charges would the 2 newly formed ions take on is Li^{+} and Br^{-}.

ycow [4]4 years ago
3 0

I believe the Lithium will give up its electron to the bromine so it would become negative

 so it is Li- and Br+

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4 0
3 years ago
How much CO2 is produced from the combustion of a tank of propane C3H8? Assume the tank contains 9kgs of propane and the molecul
Zinaida [17]

<u>Answer:</u> The mass of carbon dioxide produced is 26.9 kg

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

Given mass of propane = 9 kg = 9000 g    (Conversion factor:  1 kg = 1000 g)

Molar mass of propane = 44.1 g/mol

Putting values in equation 1, we get:

\text{Moles of propane}=\frac{9000g}{44.1g/mol}=204.08mol

The chemical equation for the combustion of propane follows:

C_3H_8+5O_2\rightarrow 3CO2+4H_2O

By Stoichiometry of the reaction:

1 mole of propane produces 3 moles of carbon dioxide.

So, 204.08 moles of propane will produce = \frac{3}{1}\times 204.08=612.24mol of carbon dioxide.

Now, calculating the mass of carbon dioxide by using equation 1, we get:

Molar mass of carbon dioxide = 44.01 g/mol

Moles of carbon dioxide = 612.24 moles

Putting values in equation 1, we get:

612.24mol=\frac{\text{Mass of carbon dioxide}}{44.01g/mol}\\\\\text{Mass of carbon dioxide}=(612.24mol\times 44.01g/mol)=26944.7g=26.9kg

Hence, the mass of carbon dioxide produced is 26.9 kg

3 0
3 years ago
Which displacement reaction is most likely to occur? Use the resources to help you. A. BaCl2 + Ca → CaCl2 + Ba B. BaCl2 + Sr → S
umka2103 [35]

Correct part is C.

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The metal which lies below in the reactivity series CANNOT displace the metal which lies above it. Similarly if the metal lies above in the reactivity series, it CAN displace the metal lying low in the series.

In A part, the reaction follows

BaCl_2+ Ca\rightarrow CaCl_2+Ba

Here Ca metal lies low in the series than Ba, therefore it cannot displace Ba metal in the reaction.

In B part ,the reaction follows

BaCl_2+Sr\rightarrow SrCl_2+Ba

Here also Sr metal lies low in the series, therefore it cannot displace Ba.

In C part, the reaction follows

BaCl_2+2Na\rightarrow 2NaCl+Ba

Here Na metal lies above in the series than Ba, therefore it can easily displace Ba in the reaction and will lead to form NaCl as a product.

In D part, the reaction follows

BaCl_2+Mg\rightarrow MgCl_2+Ba

Here Mg lies low in the series, therefore it cannot displace Ba in the reaction.

You can use the image attached.

6 0
4 years ago
Use the successive ionization energies for this unknown element to identify the family it belongs to.
NNADVOKAT [17]

Answer: Belongs to the group 2A

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As you can see, the first two ionization energies are close and low, meaning that this element ionizates easily.

Not only loses easily the first electron, but the second too

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Now, elements easily ionizable are the ones from group IA, group 2A and transition metals.

The last ones have mixed characteristics in matter of how many electrons you can remove from them, so they are not a family.

Now the question: group I or group II ?

The elements of group I have low  ionization energies  for the first electron but high energies for the second ones.

Being all that said, the unknown element belongs to the Group 2A

4 0
3 years ago
[1 mark]
andrey2020 [161]
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6 0
3 years ago
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