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3 years ago

Find the final product for: Li + Br and show work.

Chemistry

1 answer:

miskamm [114]3 years ago

7 0

Lithium is atomic number 3, so it has valency 1

While, Bromine is atomic number 35, and has valency 1

Lithium has an extra electron while Bromine need an electron, since they both need and have one electron, the form

LiBr (Lithium Bromide) where Li is +ve charged and Br is -ve charged

Happy to help :)

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In a series circuit, electrons only flow along one path. Given this information, which of the following is an advantage of serie

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Answer: D. Series circuits are simpler than parallel circuits.

Explanation:

In series circuit, all the components are connected in single path of current flow.

If one part of the circuit breaks, complete circuit would break and all the other parts also would not operate.

The voltage is not same everywhere. It is different across each component.

In series circuit, net resistance is more as resistance of all the components connected in series are added up.

So, the only advantage of a series circuit is that is simpler than the parallel circuits. Thus, the correct option is D.

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3 years ago

B. if 6.73 g of na2co3 is dissolved in enough water to make 250. ml of solution, what is the molar concentration of sodium carbo

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Answer:- $[Na_2CO_3]=0.254M$ , $[Na^+]=0.508 M$ , $[CO_3^2^-]=0.254M$

Solution:- We are asked to calculate the molarity of sodium carbonate solution as well as the sodium and carbonate ions.

Molarity is moles of solute per liter of solution. We have been given with 6.73 grams of sodium carbonate and the volume of solution is 250.mL. Grams are converted to moles and mL are converted to L and finally the moles are divided by liters to get the molarity of sodium carbonate.

Molar mass of sodium carbonate is 105.99 gram per mol. The calculations for the molarity of sodium carbonate are shown below:

$\frac{6.73gNa_2CO_3}{250.mL}(\frac{1mol}{105.99g})(\frac{1000mL}{1L})$

= $0.254MNa_2CO_3$

So, molarity of sodium carbonate solution is 0.254 M.

sodium carbonate dissociate to give the ions as:

$Na_2CO_3(aq)\rightarrow 2Na^+(aq)+CO_3^2^-$

There is 1:2 mol ratio between sodium carbonate and sodium ion. So, the molarity of sodium ion will be two times of sodium carbonate molarity.

$[Na^+]=2(0.254M)$ = 0.508 M

There is 1:1 mol ratio between sodium carbonate and carbonate ion. So, the molarity of carbonate ion will be equal to the molarity of sodium carbonate.

$[CO_3^2^-]=0.254M$

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Answer:

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