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Alenkinab [10]
2 years ago
7

Al comenzar la reacción: N2(g) + 2O2(g) ------> 2NO2(g) existe 1 mol de N2 y 2 moles de O2 y al

Chemistry
1 answer:
miss Akunina [59]2 years ago
7 0

Answer:

65

Explanation:

estion

Al comenzar la reacción: N2(g) + 2O2(g) ------> 2NO2(g) existe 1 mol de N2 y 2 moles de O2 y al

finalizarla está presente una mezcla formada por 2,2 moles en total, ¿cuál es el rendimiento para la

reacción?

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Which substance is liquid at 0°C? 1. alcohol 2. aluminum 3. ammonia 4. copper
Tanya [424]
Alcohol, alcohol is the only liquid  since copper and aluminum are a solid ammonia is a bacteria
5 0
3 years ago
How many moles of nitric acid are present in 35.0 ml of a 2.20 M solution?
vodomira [7]

Answer:

There are 77 millimoles of nitric acid present in 35.0 mL of a 2.20 M solution

Explanation:

Molarity of the solution = 2.20 M

Molarity=\frac{number\:of\:moles}{Volume\:of\:Solution\:in\:L}\\\\Number\:of\:moles=Molarity\times(Volume\:of\:Solution\:in\:L)\\\\Volume\:of\:Solution=35\:mL=35\times10^{-3}L\\\\Number\:of\:moles=2.20\times35\times10^{-3}=77\times10^{-3}\:moles\:of\:HNO_{3}

Therefore, there are 77 millimoles of nitric acid present in 35.0 mL of a 2.20 M solution

8 0
3 years ago
Identify a process that is NOT reversible. A. melting of steel B. freezing water C. melting of ice D. frying an egg E. depositio
vredina [299]
I’m pretty sure it would be D. Frying an egg
8 0
3 years ago
Part C: complete the third column <br> Part D: complete the fourth column
Helga [31]

Answer:

Part C: P2 = 0.30 atm

Part D: V1 = 16.22 L.

Explanation:

Part C:

Initial pressure (P1) = 2.67 atm

Initial volume (V1) = 5.54 mL

Final pressure (P2) =.?

Final volume (V2) = 49 mL

The final pressure (P2) can be obtained as follow:

P1V1 = P2V2

2.67 x 5.54 = P2 x 49

Divide both side by 49

P2 = (2.67 x 5.54)/49

P2 = 0.30 atm

Therefore, the final pressure (P2) is 0.30 atm

Part D:

Initial pressure (P1) = 348 Torr

Initial volume (V1) =?

Final pressure (P2) = 684 Torr

Final volume (V2) = 8.25 L

The initial volume (V1) can be obtained as follow:

P1V1 = P2V2

348 x V1 = 684 x 8.25

Divide both side by 348

V1 = (684 x 8.25)/348

V1 = 16.22 L

Therefore, the initial volume (V1) is 16.22 L

6 0
3 years ago
How many milliliters of 0.260 m na2s are needed to react with 35.00 ml of 0.315 m agno3?
allochka39001 [22]

The complete balanced chemical reaction is:

2 AgNO3 + Na2S --> 2 NaNO3 + Ag2S

 

First let us calculate the number of moles of AgNO3.

moles AgNO3 = 0.315 M * 0.035 L

moles AgNO3 = 0.011025 mol

 

From the reaction, 1 mole of Na2S is needed for every 2 moles of AgNO3 hence:

moles Na2S required = 0.011025 mol AgNO3 * (1 mol Na2S / 2 mol AgNO3)

moles Na2S required = 5.5125 x 10^-3 mol

 

Therefore volume required is:

volume Na2S = 5.5125 x 10^-3 mol / 0.260 M

<span>volume Na2S = 0.0212 L = 21.2 mL</span>

6 0
3 years ago
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