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Alik [6]
4 years ago
7

When I2 and FeCl2 are mixed together, iodine (I) cannot replace chlorine (Cl) in the compound because iodine is lower on the per

iodic table.
True
False
Chemistry
1 answer:
Nata [24]4 years ago
3 0

Answer: The given statement is true.

Explanation: If this reaction would have occurred, then this reaction would be considered as displacement reaction.

Displacement reactions are the reaction in which more reactive element displaces the less reactive element in a chemical reaction. This is based on the reactivity of elements.

Reactivity of elements is the tendency of the elements to gain or loose electrons. The reactivity decreases down the group in a periodic table.

In the given reaction, Iodine and chlorine are the elements of the same group in the periodic table and iodine lies below chlorine in the group. So, the reactivity of iodine is less than the reactivity of chlorine.

Hence, in the given reaction, iodine will not replace chlorine because it lies below in the periodic table.

FeCl_2+I_2\rightarrow \text{No reaction}

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egoroff_w [7]

Answer:

The concentration of Cd2+ is 0.0175 M

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Explanation:

<u>Step 1:</u> Data given

A 50.0 mL sample contains Cd2+ and Mn2+

volume of 0.05 M EDTA = 56.3 mL

Titration of the excess unreacted EDTA required 13.4 mL of 0.0310 M Ca2+.

Titration of the newly freed EDTA required 28.2 mL of 0.0310 M Ca2+

<u>Step 2:</u> Calculate mole ratio

The reaction of EDTA and any metal ion is 1:1, the number of mol of Cd2+ and Mn2+  in the mixture equals the total number of mol of EDTA minus the number of mol of EDTA consumed  in the back titration with Ca2+:

<u>Step 3: </u>Calculate total mol of EDTA

Total EDTA = (56.3 mL EDTA)(0.0500 M EDTA) = 0.002815 mol EDTA

Consumed EDTA = 0.002815 mol – (13.4 mL Ca2+)(0.0310 M Ca2+) = 0.002815 - 0.0004154 = 0.0023996 mol EDTA

<u>Step 4:</u> Calculate total moles of CD2+ and Mn2+

So, the total moles of Cd2+ and Mn2+ must be 0.0023996 mol

<u>Step 5:</u> Calculate remaining moles of Cd2+

The quantity of cadmium must be the same as the quantity of EDTA freed after the reaction  with cyanide:

Moles Cd2+ = (28.2 mL Ca2+)(0.0310 M Ca2+) = 0.0008742 mol Cd2+.

<u>Step 6:</u> Calculate remaining moles of Mn2+

The remaining moles must be Mn2+: 0.0023996 - 0.0008742 = 0.0015254 moles Mn2+

<u>Step 7: </u>Calculate initial concentrations

The initial concentrations must have been:

(0.0008742 mol Cd2+)/(50.0 mL) = 0.0175 M Cd2+

(0.0015254 mol Mn2+)/(50.0 mL) = 0.0305 M Mn2+

The concentration of Cd2+ is 0.0175 M

The concentration of Mn2+ is 0.0305 M

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Answer:

check out this website,it might help.

Explanation:

https://courses.lumenlearning.com/boundless-chemistry/chapter/physical-and-chemical-properties-of-matter/

Im really not the smartest in the world.

<em><u>Website was not made by me, im just a lazy potat0.</u></em>

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