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Mnenie [13.5K]
2 years ago
13

The elements that touch the zigzag line are classified as

Chemistry
1 answer:
svp [43]2 years ago
5 0

On the periodic table everything to the right of the line is a non metal

Everything to the left is a metal. So the line crosses to the right is NM to left is M

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Between 2014 and 2016, more than 25,000 children in Flint, Michigan, drank water that was contaminated with lead from lead pipes
-Dominant- [34]

Answer:

Regularly test the water in residents' homes.

Explanation:

The only way to know if tap water contains lead is to do tests to determine the levels of that metal in the water. Therefore, the state is under an obligation to constantly conduct such tests in the resident´s homes and thus determine whether the water supplied is fit for human consumption.  

The state after the tests must guarantee the population the treatment of the water to reduce the levels of lead. The main pipes that contain lead pipes must be changed, as well as those parts of the service connections made of lead.

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a scientist finds an organism that has a single cell without a nucleus.This organism was found in pond water.In which KINGDOM do
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<em>Unicellular organism</em>

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8 0
2 years ago
Solid NaBr is slowly added to a solution that is 0.073 M in Cu+ and 0.073 M in Ag+.Which compound will begin to precipitate firs
saul85 [17]

Answer :

AgBr should precipitate first.

The concentration of Ag^+ when CuBr just begins to precipitate is, 1.34\times 10^{-6}M

Percent of Ag^+ remains is, 0.0018 %

Explanation :

K_{sp} for CuBr is 4.2\times 10^{-8}

K_{sp} for AgBr is 7.7\times 10^{-13}

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgBr has a smaller than CuBr then AgBr should precipitate first.

Now we have to calculate the concentration of bromide ion.

The solubility equilibrium reaction will be:

CuBr\rightleftharpoons Cu^++Br^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Cu^+][Br^-]

4.2\times 10^{-8}=0.073\times [Br^-]

[Br^-]=5.75\times 10^{-7}M

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

AgBr\rightleftharpoons Ag^++Br^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Ag^+][Br^-]

7.7\times 10^{-13}=[Ag^+]\times 5.75\times 10^{-7}M

[Ag^+]=1.34\times 10^{-6}M

Now we have to calculate the percent of Ag^+ remains in solution at this point.

Percent of Ag^+ remains = \frac{1.34\times 10^{-6}}{0.073}\times 100

Percent of Ag^+ remains = 0.0018 %

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The answer would be Force
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Answer: haii~! ur answer is A.) true

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