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zhannawk [14.2K]

3 years ago

12

Scoring Scheme: 3-3-2-1 Given the starting solution had a concentration of 1.25 M, how many moles of Co[H2O]6Cl2 were available

in the amount of starting solution you used

1 answer:

Pepsi [2]3 years ago

8 0

Answer:

The solution has 0.00994 moles of Co[H2O]6Cl2

Explanation:

Complete question: Given the starting solution had a concentration of 1.25 M, how many moles of Co[H2O]6Cl2 were available in the amount of starting solution you used?

mL= 7.95

Step 1: Data given

Concentration of the starting solution = 1.25 M

Starting solution = Co[H2O]6Cl2

Step 2: Calculate molar mass of Co[H2O]6Cl2

Atomic mass of Co = 58.93 g/mol

Atomic mass of H = 1.01 g/mol

Atomic mass of O = 16.00 g/mol

atomic mass of Cl = 35.45 g/mol

Molar mass of Co[H2O]6Cl2 =

1*58.93 + 12*1.01 + 6*16.00 + 2*35.45 = 237.95 g/mol

Step 3: Calculate number of moles

C = n/v

⇒with C = the Concentration of the starting solution = 1.25 M

⇒with n= the number of moles = to be determined

⇒with v = the volume = 7.95 mL = 0.00795 L

n = C * V

n = 1.25 M * 0.00795 L

n = 0.00994 moles

The solution has 0.00994 moles of Co[H2O]6Cl2

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Answer:

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Explanation:

To predict the products of these reactions we need to know the kind of reactions. All these reactions are double replacement reaction. In these kinds of reactions, the products will be the result of exchanging ions in the reactants. So, the first step is to identify the ions.

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So, the products for this reaction will be
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After this, we proceed to balance the equation. For this, we check that we have the same number of each element on both sides of the equation. In this case, we can see that we have the same number, so the equation is balanced. Finally, we check the rules of solubility to see if the species are soluble in water or not. In this case sulfates area always soluble except for mercury so Hg2(SO4) precipitates in the solution (pre). Nitrates are always soluble so Cu(NO3)2 is soluble (aq) 
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The complete ionic equation allows to show which of the reactants or products exist primarily as ions. For this reaction this will be:

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To get net ionic equation we take away the ions that did not participate in the reactions. In other words the ones that are the same on both sides in the equation. In this case we see that [tex] Cu^{+2}(aq) and (NO_{3})^{-1}(aq) [/text] are the same on both sides so those ions are not include in the net ionic equation. This is:
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