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MissTica
3 years ago
11

Gaseous ammonia chemically reacts with oxygen O2 gas to produce nitrogen monoxide gas and water vapor. Calculate the moles of ox

ygen needed to produce 1.80mol of water
Chemistry
1 answer:
WINSTONCH [101]3 years ago
6 0

Answer : The number of moles of oxygen needed are, 1.5 moles.

Explanation :

The balanced chemical reaction will be:

4NH_3+5O_2\rightarrow 4NO+6H_2O

Now we have to calculate the moles of oxygen.

From the balanced chemical reaction we conclude that,

As, 6 moles of water vapor produces from 5 moles of oxygen

So, 1.80 moles of water vapor produces from \frac{5}{6}\times 1.80=1.5 moles of oxygen

Therefore, the number of moles of oxygen needed are, 1.5 moles.

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Answer:

i think i like 14 or 8 atom

Explanation:

8 0
3 years ago
According to reference table adv-10, which reaction will take place spontaneously?
olga_2 [115]
Missing table!! write the elements with the first letter of the symbol with Upper Caps letters!!!

http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/EStandardTable.htm

<span>Ni2+ +Pb(s) → Ni(s) + Pb2+
</span>The potential of the oxidation of Pb(s) --> Pb2+(aq) is 0.126 V 
The potential of the reduction go Ni2+(aq) --> Ni(s) is -0.25 V 

<span>Add the two together and the potential for the reaction is -0.124 V (NO SPONTANEOUS THE SIGN IS NEGATIVE)

</span><span>au3+ + al(s) → au(s) + al3+Au3+(aq) ->   Au(s)  +1.5 VAl -> Al3+  +1.66VV= 3.16 (SPONTANEOUS THE SIGN OF THE PONTENTIAL IS POSITIVE)</span><span>Sr2+ + Sn(s) → Sr(s) + Sn2+
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Sr2+(aq) + 2 e–  <span>  Sr(s)  V= -2.89V
</span>Sn -> Sn2+ V= 0.14 V
V= -2.75 V (no spontaneous)

<span>Fe2+ + Cu(s) → Fe(s) + Cu2+
</span>Fe2+(aq) + 2 e–<span>  </span><span>  Fe(s)  V= -0.44 V
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5 0
3 years ago
Why can’t you perform single displacement reactions with group IA metals?
pishuonlain [190]
Answer is: because alkaline metals (group IA metals) are the strongest reducing agents and most reactive metals.
Reducing agent<span> is an element  or compound that loses an </span>electron<span> to another </span>chemical species<span> in a </span>redox <span>chemical reaction and they have been oxidized.
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5 0
3 years ago
Be sure to answer all parts. Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after
Fantom [35]

Answer:

(a) pH = 12.73

(b) pH = 10.52

(c) pH = 1.93

Explanation:

The net balanced reaction equation is:

KOH + HBr ⇒ H₂O + KBr

The amount of KOH present is:

n = CV = (0.1000 molL⁻¹)(30.00 mL) = 3.000 mmol

(a) The amount of HBr added in 9.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(9.00 mL) = 0.900 mmol

This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:

(3.000 mmol) - (0.900 mmol) = 2.100 mmol KOH

After the addition of HBr, the volume of the KOH solution is 39.00 mL. The concentration of KOH is calculated as follows:

C = n/V = (2.100 mmol) / (39.00 mL) = 0.0538461 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0538461) = 1.2688

pH = 14 - pOH = 14 - 1.2688 = 12.73

(b) The amount of HBr added in 29.80 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(29.80 mL) = 2.980 mmol

This amount of HBr will neutralize an equivalent amount of KOH, leaving the following amount of KOH:

(3.000 mmol) - (2.980 mmol) = 0.0200 mmol KOH

After the addition of HBr, the volume of the KOH solution is 59.80 mL. The concentration of KOH is calculated as follows:

C = n/V = (0.0200 mmol) / (59.80 mL) = 0.0003344 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0003344) = 3.476

pH = 14 - pOH = 14 - 3.476 = 10.52

(c) The amount of HBr added in 38.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(38.00 mL) = 3.800 mmol

This amount of HBr will neutralize all of the KOH present. The amount of HBr in excess is:

(3.800 mmol) - (3.000 mmol) = 0.800 mmol HBr

After the addition of HBr, the volume of the analyte solution is 68.00 mL. The concentration of HBr is calculated as follows:

C = n/V = (0.800 mmol) / (68.00 mL) = 0.01176 M HBr

The pH of the solution can then be calculated:

pH = -log[H⁺] = -log(0.01176) = 1.93

4 0
3 years ago
(6.022*10^23)*2.58=<br> what’s the answer
Sladkaya [172]
The answer would be
1.553676 \times  {10}^{24}
8 0
3 years ago
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