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DENIUS [597]
2 years ago
9

Ammonia, NH3(g), can be used as a substitute for fossil fuels in some internal combustion

Chemistry
1 answer:
sammy [17]2 years ago
7 0

Answer:

4 NH₃ (g) + 3 O₂ (g) ----> 2 N₂ (g) + 6 H₂O (g) + energy

Explanation:

The unbalanced equation:

NH₃ (g) + O₂ (g) ----> N₂ (g) + H₂O (g) + energy

<u>Reactants:</u> 1 nitrogen, 3 hydrogen, 2 oxygen

<u>Products:</u> 2 nitrogen, 2 hydrogen, 1 oxygen

The sort-of balanced equation

2 NH₃ (g) + 1/2 O₂ (g) ----> N₂ (g) + 3 H₂O (g) + energy

I first started by balancing the nitrogens. This then lead me to balance the hydrogens. When I went to balance the oxygens, only a 1/2 coefficient would work. However, 1/2 coefficients are technically incorrect. To get rid of this coefficient, you need to multiply each coefficient by 2.

The balanced equation:

4 NH₃ (g) + 3 O₂ (g) ----> 2 N₂ (g) + 6 H₂O (g) + energy

<u>Reactants:</u> 4 nitrogen, 12 hydrogen, 6 oxygen

<u>Products:</u> 4 nitrogen, 12 hydrogen, 6 oxygen

Now, all of the elements are equal on both sides and there are no 1/2 coefficients. These are the smallest whole-number coefficients because some of the numbers cannot be further simplified (3).

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An intravenous infusion is to contain 15 mEq of potassium ion and 20 mEq of sodium ion in 500 mL of 5% dextrose injection. Using
Studentka2010 [4]

Answer:

To supply the required ions it is necessary to inject 5,6mL of 6g/30mL solution and 131,1 mL of 0,9% solution.

Explanation:

1mEq of sodium are 59mg of NaCl and 1mEq of potassium are 75mg KCl

in intravenous infusion 15 mEq of K are:

15x75mg KCl = 1,125g of KCl

And 20 mEq of Na are:

20x59mg NaCl = 1,18g of NaCl

To supply the potassium ion it is necessary to inject:

1,125g of KCl×\frac{30mL}{6g} =<em> 5,6mL  of 6g/30mL solution</em>

And, to supply the sodium ion it is necessary to inject:

1,18g of NaCl×\frac{100mL}{0,9g} = <em>131,1 mL of 0,9% solution</em>

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I hope it helps!

6 0
3 years ago
Which of the following is NOT a product obtained from the distillation of coal tar?
rjkz [21]

Answer:

Coke.

Explanation:

The distillation of coal tar can obtain aromatic compounds like benzene and toluene, and also phenolic compounds like phenol. Hence, the only option here that cannot be obtained from the distillation of coal tar is coke.

Hope this helped!

7 0
3 years ago
Read 2 more answers
If a proton and electron were 0.40nm apart, predict what the force of attraction might be
Aleks [24]
??????????????????????????????????????????????                              
3 0
2 years ago
Which of these is most likely to create large areas of land subsidence?
Mumz [18]
I think it is Global warming
6 0
2 years ago
In acidic solution, the sulfate ion can be used to react with a number of metal ions. One such reaction is SO42−(aq)+Sn2+(aq)→H2
allsm [11]

Answer:

The final balanced equation is :

SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)

Explanation:

SO_4^{2-}(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+Sn^{4+}(aq)

Balancing in acidic medium:

First we will determine the oxidation and reduction reaction from the givne reaction :

Oxidation:

Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)

Balance the charge by adding 2 electrons on product side:

Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)+2e^-....[1]

Reduction :

SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)

Balance O by adding water on required side:

SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)

Now, balance H by adding H^+ on the required side:

SO_4^{2-}(aq)+4H^+(aq)\rightarrow H_2SO_3(aq)+H_2O(l)

At last balance the charge by adding electrons on the side where positive charge is more:

SO_4^{2-}(aq)+4H^+(aq)+2e^-\rightarrow H_2SO_3(aq)+H_2O(l)..[2]

Adding [1] and [2]:

SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)

The final balanced equation is :

SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)

4 0
3 years ago
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