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oksian1 [2.3K]
2 years ago
11

Given the chemical formulas of the following compounds, name each compound and state the rules you used to determine each name.

a. K2S b. FeO c. NH4OH
Chemistry
1 answer:
Ede4ka [16]2 years ago
6 0

Answer:

1.potassium sulphide 2. iron oxide 3. ammonium hydroxide

Explanation:

1. K2S has k+ ion and s2- ion which is known as sulphide ion therefore the name becomes potassium sulfide.

2. FeO has Fe2+ ion and O2- ion which is oxide ion therefore the name becomes Iron Oxide.

3. NH4OH has NH4+ ion hich is ammonium ion and OH- which is hydroxide ion therefore the name becomes ammonium hydroxide.

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What do the three types of mixtures have in common? <br> (colloid, solution, suspension)
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Answer:

A colloid is a heterogeneous mixture in which the dispersed particles are intermediate in size between those of a solution and a suspension. The particles are spread evenly throughout the dispersion medium, which can be a solid, liquid, or gas.

Explanation:

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Draw the Lewis structure, including unshared pairs, of the following molecules. Carbon has four bonds in the compound.
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Explanation:

Refer to attachment.

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A solid powder is composed of molecules containing silver (Ag), nitrogen (N), and oxygen (O) atoms. All of the molecules are ide
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Answer is in the file below

tinyurl.com/wpazsebu

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2 years ago
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A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 64.0 mL of 0.0600 M EDTA . Titration of the excess unreacted EDTA req
tigry1 [53]

Answer:

the concentration of Cd^{2+}  in the original solution= 0.0088 M

the concentration of Mn^{2+} in the original solution = 0.058 M

Explanation:

Given that:

The volume of the sample  containing Cd2+ and Mn2+ =  50.0 mL; &

was treated with 64.0 mL of 0.0600 M EDTA

Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+

i.e the strength of the Ca2+ = 0.0310 M

Titration of the newly freed EDTA required 14.2 mL of 0.0310 M Ca2+

To determine the concentrations of Cd2+ and Mn2+ in the original solution; we have the following :

Volume of newly freed EDTA = \frac{Volume\ of \ Ca^{2+}* Sample \ of \ strength }{Strength \ of EDTA}

= \frac{14.2*0.0310}{0.0600}

= 7.3367 mL

concentration of  Cd^{2+} = \frac{volume \ of \  newly  \ freed \ EDTA * strength \ of \ EDTA }{volume \ of \ sample}

= \frac{7.3367*0.0600}{50}

= 0.0088 M

Thus the concentration of Cd^{2+} in the original solution = 0.0088 M

Volume of excess unreacted EDTA = \frac{volume \ of \ Ca^{2+} \ * strength \ of Ca^{2+} }{Strength \ of \ EDTA}

= \frac{16.1*0.0310}{0.0600}

= 8.318 mL

Volume of EDTA required for sample containing Cd^{2+}   and  Mn^{2+}  = (64.0 - 8.318) mL

= 55.682 mL

Volume of EDTA required for Mn^{2+}  = Volume of EDTA required for

                                                                sample containing  Cd^{2+}   and  

                                                             Mn^{2+} --  Volume of newly freed EDTA

Volume of EDTA required for Mn^{2+}  = 55.682 - 7.3367

= 48.3453 mL

Concentration  of Mn^{2+} = \frac{Volume \ of EDTA \ required \ for Mn^{2+} * strength \ of \ EDTA}{volume \ of \ sample}

Concentration  of Mn^{2+} =  \frac{48.3453*0.0600}{50}

Concentration  of Mn^{2+}  in the original solution=   0.058 M

Thus the concentration of Mn^{2+} = 0.058 M

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Jim went for a 3.75 hour run. How many minutes did he run?
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