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Dimas [21]
2 years ago
12

Ammonium phosphate is an important ingredient in many solid fertilizers. It can be made by reacting aqueous phosphoric acid with

liquid ammonia. Calculate the moles of ammonium phosphate produced by the reaction of of ammonia. Be sure your answer has a unit symbol, if necessary, and round it to significant digits.
Chemistry
1 answer:
Paladinen [302]2 years ago
5 0

The given question is incomplete. The complete question is:

Ammonium phosphate (NH_4)_3PO_4 is an important ingredient in many solid fertilizers. It can be made by reacting aqueous phosphoric acid H_3PO_4 with liquid ammonia. Calculate the moles of ammonium phosphate produced by the reaction of 1.7 mol of ammonia. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.  

Answer: 0.57 moles of ammonium phosphate are produced by the reaction of 1.7 moles of liquid ammonia

Explanation:

The balanced chemical equation is:

H_3PO_4+3NH_4OH\rightarrow (NH_4)_3PO_4+3H_2O

According to stoichiometry:

3 moles of liquid ammonia (NH_4OH) produces = 1 mole of ammonium phosphate (NH_4)_3PO_4

Thus 1.7 moles of liquid ammonia (NH_4OH) produces = \frac{1}{3}\times 1.7=0.57 mole of ammonium phosphate (NH_4)_3PO_4

Thus 0.57 moles of ammonium phosphate are produced by the reaction of 1.7 moles of liquid ammonia

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An electrochemical cell at 25°C is composed of pure copper and pure lead solutions immersed in their respective ionis. For a 0.6
ExtremeBDS [4]

Answer :

(a) The concentration of Pb^{2+} is, 0.0337 M

(b) The concentration of Pb^{2+} is, 6.093\times 10^{32}M

Solution :

<u>(a) As per question, lead is oxidized and copper is reduced.</u>

The oxidation-reduction half cell reaction will be,

Oxidation half reaction:  Pb\rightarrow Pb^{2+}+2e^-

Reduction half reaction:  Cu^{2+}+2e^-\rightarrow Cu

The balanced cell reaction will be,  

Pb(s)+Cu^{2+}(aq)\rightarrow Pb^{2+}(aq)+Cu(s)

Here lead (Pb) undergoes oxidation by loss of electrons, thus act as anode. Copper (Cu) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

E^o_{[Pb^{2+}/Pb]}=-0.13V

E^o_{[Cu^{2+}/Cu]}=+0.34V

E^o=E^o_{[Cu^{2+}/Cu]}-E^o_{[Pb^{2+}/Pb]}

E^o=0.34V-(-0.13V)=0.47V

Now we have to calculate the concentration of Pb^{2+}.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Pb^{2+}]}{[Cu^{2+}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = 0.507 V

Now put all the given values in the above equation, we get:

0.507=0.47-\frac{0.0592}{2}\log \frac{[Pb^{2+}]}{(0.6)}

[Pb^{2+}]=0.0337M

Therefore, the concentration of Pb^{2+} is, 0.0337 M

<u>(b) As per question, lead is reduced and copper is oxidized.</u>

The oxidation-reduction half cell reaction will be,

Oxidation half reaction:  Cu\rightarrow Cu^{2+}+2e^-

Reduction half reaction:  Pb^{2+}+2e^-\rightarrow Pb

The balanced cell reaction will be,  

Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)

Here Copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

E^o_{[Pb^{2+}/Pb]}=-0.13V

E^o_{[Cu^{2+}/Cu]}=+0.34V

E^o=E^o_{[Pb^{2+}/Pb]}-E^o_{[Cu^{2+}/Cu]}

E^o=-0.13V-(0.34V)=-0.47V

Now we have to calculate the concentration of Pb^{2+}.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}]}{[Pb^{2+}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = 0.507 V

Now put all the given values in the above equation, we get:

0.507=-0.47-\frac{0.0592}{2}\log \frac{(0.6)}{[Pb^{2+}]}

[Pb^{2+}]=6.093\times 10^{32}M

Therefore, the concentration of Pb^{2+} is, 6.093\times 10^{32}M

6 0
3 years ago
Ice cream is made by freezing a liquid mixture that, as a first approximation, can be considered a solution of sucrose (C12H22O1
AlladinOne [14]

Answer:

Hi.

The temperature is approximately zero degrees (0°C)

Explanation:

It is important to keep in mind that in the production of ice cream the decrease in the freezing point of the water present in the mixture is called the antifreeze power of the mixture. In ice cream, the freezing point decrease will be caused by each substance that is dissolved in the mixture: lactose, salts, sugars and any other substance. Each of these substances will contribute to the decrease in the freezing point of the mixture. The phase diagram attached in the file shows the sugar solutions in water. When a solution cools (point A), there comes a time when the freezing curve is reached (point B). At that moment ice begins to appear. As shown in the diagram this temperature is approximately zero degrees (0 ° C).

7 0
3 years ago
A student submits the following work on the mass number, but he has made a few mistakes. Select the sentences that are incorrect
rewona [7]

Answer

 

Calculating the mass number for an atom requires that we know the atomic number and the number of protons in the atom’s nucleus. The mass number then gives us the average weight of atoms of a given element. However, as long as the number of protons equals the number of neutrons, the values balance out and we always obtain a whole number for the mass number.

Explanation:

those 3

7 0
3 years ago
Consider the first-order reaction described by the equation At a certain temperature, the rate constant for this reaction is 5.8
zubka84 [21]

<u>Answer:</u> The half life of the reaction is 1190.7 seconds

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

t_{1/2}=\frac{0.693}{k}

where,

k = rate constant of the reaction = 5.82\times 10^{-4}s^{-1}

t_{1/2} = half life of the reaction = ?

Putting values in above equation, we get:

t_{1/2}=\frac{0.693}{5.82\times 10^{-4}s^{-1}}=1190.7s

Hence, the half life of the reaction is 1190.7 seconds

8 0
3 years ago
How many grams of Chromium can be formed when 150 grams of
valentinak56 [21]

Answer: 104 g

Explanation: reaction Cr2O3 + 3 H2 ⇒ 2 Cr + 3 H2O

M(Cr2O3) = 150 g/mol, so n = m/M = 1.0 mol

Number of moles of H2 should be 3.0 moles and

It is much greater (150 g / 2.016 g/mol)

1 mol Cr2O3 produces 2 mol Cr.

Mass m= 2.0 mol· 52g/mol= 104 g

6 0
2 years ago
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