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Lelechka [254]
3 years ago
8

Give the name and formula for the acid derived from each of the following anions: (a) perchlorate (b) N03- (c) bromite (d) H2PO4

-
Chemistry
1 answer:
Vlada [557]3 years ago
6 0

Answer:

(a) Perchloric acid (HClO₄)

(b) Nitric acid (HNO₃)

(c) Bromous acid (HBrO₂)

(d) Phosphoric Acid (H₃PO₄)

Explanation:

Acids are the species which furnish hydrogen ions in the solution. The other specie produced with the hydrogen ion is called its conjugate base.

Thus,

(a) Perchlorate (ClO₄⁻) ion:

The acid that furnish perchlorate ion is <u>perchloric acid (HClO₄)</u>.

(b)  NO₃⁻ ion:

The acid that furnish NO₃⁻ ions is <u>Nitric acid (HNO₃)</u>.

(c) bromite (BrO₂⁻) ion:

The acid that furnish BrO₂⁻ ions is <u>Bromous acid (HBrO₂)</u>.

(d)  H₂PO₄⁻ ion:

The acid that furnish H₂PO₄⁻ ions is <u>Phosphoric Acid (H₃PO₄)</u>.

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Answers:

Question 1:

The diagram for gallium will have flat, horizontal lines at <u><em>30 and 2204</em></u><em> </em>°C.

Questoin 2:

The diagram for methane will have a <u><em>diagonal </em></u>line representing the <u><em>liquid phase</em></u> between -183°C and -162°C.

Question 3:

For gold, the boiling point corresponds to the y-value at <u><em>2856</em></u><u> </u>°C of <u><em>the top horizontal line</em></u>

Question 4.

For nitrogen, the line at -210°C will be<u><em> flat</em></u>, which represents <u><em>the change from a solid to a liquid</em></u>

Explanations:

Question 1:

The diagram for gallium will have flat, horizontal lines at <u><em>30 and 2204</em></u><em> </em>°C.

The table shows that the melting point of gallium is 30°C and its boiling point is 2204°C.

<em>Melting point</em> is the temperature at which the substace changes its state from<em> solid to liquid</em>. During that change, <em>the temperature</em> of the substance <em>does not change</em>, because the heat supplied is used to accomplish the phase change. So, the temperature is constant and that means <em>that portion of the diagram is flat</em>.

The same is valid during<em> boiling</em>: the temperature remains constant while the substance is passing<em> from liquid to gas</em> at the boiling point.

Questoin 2:

The diagram for methane will have a <u><em>diagonal </em></u>line representing the <u><em>liquid phase</em></u> between -183°C and -162°C.

Between the <em>melting</em> (-183°C) and<em> boiling</em> (-162°C) points of methane, its temperature will increase more or less linearly, which is represented with a <em>diagonal</em> (slant) <em>line</em> between those points. During this interval the heat is used to <em>increase the temperature</em> and no phase of change happens.

Question 3:

For gold, the boiling point corresponds to the y-value at <u><em>2856</em></u><u> </u>°C of <u><em>the top horizontal line</em></u>

<u><em></em></u>

The table shows that the<em> boiling point</em> of gold is 2,856°C.

In a <em>temperature-vs.-time diagram</em> the<em> temperature is represented on the vertical axis (y-value)</em> and the time is represented on the horizontal axis.

Since, the temperature of the substance does not change during <em>boiling,</em> the line during the time that this change of phase is happening is flat. And since this temperatue is higher than the melting temperature, this is the <em>top horizontal line in the diagram</em>.

Question 4.

For nitrogen, the line at -210°C will be<u><em> flat</em></u>, which represents <u><em>the change from a solid to a liquid</em></u>

<u><em></em></u>

The table shows that the <em>melting point </em>of nitrogen is -210°C, that means that the temperature will remain constant at -210°C while the substance is absorbing heat to pass from solid to liquid.

<u>In conclusion, you must remember that all the phase changes, melting (from solid to liquid), freezing (from liquid to solid), boilng (from liquid to gas), and condensing (from gas to liquid) happens at constant temperature, and so the </u><em><u>temperature - vs. - time diagrams </u></em><u>show flat lines (constant y-values) during those intervals of time.</u>

4 0
2 years ago
Read 2 more answers
Consider the reaction 2 al + Fe2O3 to 2Fe + Al2O3. If 60.0g of Al is reacted with excess Fe2O3, determine the amount (in moles)
olganol [36]

 The  amount  of  Al2O3  in moles=  1.11 moles    while in  grams   = 113.22 grams


    <em><u>calculation</u></em>

     2 Al  + Fe2O3 → 2Fe  + Al2O3

    step  1: find the moles of Al  by  use of <u><em>moles= mass/molar  mass  </em></u>formula

    =  60.0/27= 2.22  moles


    Step 2: use the mole ratio to determine the  moles of Al2O3.

 The  mole ratio  of Al : Al2O3 is  2: 1 therefore the moles of Al2O3= 2.22/2=1.11  moles


Step 3:    finds the mass  of  Al2O3  by us of  <u><em>mass= moles x molar mass</em></u><em> </em>formula.

The molar  mass of Al2O3  =  (2x27)  +( 16 x3) = 102  g/mol

mass is therefore=  102  g/mol  x 1.11= 113.22 grams


             

7 0
2 years ago
Use the following half-reactions to write three spontaneous reactions, calculate E°cell for each reaction, and rank the oxidizin
Ainat [17]

Answer:

See explaination

Explanation:

1)

we know that

half cell with higher reduction potential is cathode

so

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

anode :

Cr(s) ---> Cr+3 + 3e-

so

overall reaction is

3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3

now

Eo cell = Eo cathode - Eo anode

so

EO cell = 1.77 + 0.74

Eo cell = 2.51 V

now

in this case

oxidizing agents are N20 and Cr+3

reducing agents are Cr and N2

higher the reduction potential , stronger the oxidizing agent

lower the reduction potential , stronger the reducing agent

so

oxidzing agents

N20 > Cr+3

reducing agents

Cr > N2

2)

cathode :

Au+ + e- --> Au

anode :

Cr ---> Cr+3 + 3e-

overall reaction

3Au+ + Cr ---> 3Au + Cr+3

Eo cell = 1.69 + 0.74

Eo cell = 2.43

now

oxidizing agents :

Au+ > Cr+3

reducing agents :

Cr > Au

3)

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

andoe :

Au ---> Au+ + e-

overall

2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20

Eo cell = 1.77 - 1.69

Eo cell = 0.08

oxidizing agents

N20 > Au+

reducing agents

Au > N2

8 0
3 years ago
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