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Bad White [126]
3 years ago
9

If 16.00 g of O₂ reacts with 80.00 g NO, what is the excess reactant? * A.O₂ B.NO

Chemistry
1 answer:
Fed [463]3 years ago
8 0

Answer:

  B. NO

Explanation:

In each mole of NO, the weight of the oxygen is 16.00 g; the weight of the nitrogen is 14.01 g. Then in 80 g of NO, we have 42.65 g of oxygen and 37.35 g of nitrogen.

__

In each mole of NO₂, the weight of the oxygen is 2(16.00) = 32 g. The weight of the nitrogen is 1(14.01) = 14.01 g.

From the available oxygen, we can produce ...

 (42.65 g +16 g)/(32.00 g/mol) = 1.83 mol of NO₂

From the available nitrogen, we can produce ...

 (37.35 g)/(14.01 g/mol) = 2.67 mol of NO₂

Clearly, the reaction is limited by the amount of available oxygen.

NO is the excess reactant.

__

Read more at ...

brainly.com/question/16500145

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Replication mistakes happen. Sometimes DNA polymerase enzymes add the incorrect nucleotide, too many, or too few nucleotides to a sequence. Or the DNA polymerase gets looped on the same strand, adding a few codons in a repetition before the strand slips away and it continues, or there is a tangle of crossing DNA and it jumps to the incorrect thread and back (or never), etc. In order to ensure that the bases added to a developing strand are appropriately matched with their complements, DNA polymerase enzymes are quite picky about the nucleotides they choose to use. However, these enzymes do make errors. Specifically, at a rate of around 1 per 100,000 nucleotides. Doesn't sound like much, but since each diploid cell has 6 billion base pairs, there are around 120,000 errors every cell division.

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1) The densities of air at −85°C, 0°C, and 100°C are 1.877 g dm−3, 1.294 g dm−3, and 0.946 g dm−3, respectively. From these data
earnstyle [38]

Answer:

1) The absolute zero temperature is -272.74 °C

2) The absolute zero temperature is -269.91°C

Explanation:

1) The specific volume of air at the given temperatures are;

At -85°C, v =  1/(1.877) = 0.533 dm³/g, the pressure =

At 0°C, v = 1/1.294 ≈ 0.773 dm³/g

At 100°C, v = 1/0.946 ≈ 1.057 dm³/g

We therefore have;

v₁ = 0.533 T₁ = 188.15  K

v₂ = 0.773 T₂ = 273.15  K

v₃ = 1.057 T₃ = 373.15  K

The slope of the graph formed by the above data is therefore given as follows;

m = (1.057 - 0.533)/(373.15 - 188.5) = 0.00284 dm³/K

The equation is therefor;

v - 0.533 = 0.00284×(T - 188.15)

v = 0.00284×T - 0.534 + 0.533  = 0.00284×T - 0.001167

v =  0.00284×T - 0.001167

Therefore, when the temperature, at absolute 0, we have;

v = 0

Which gives;

0 =  0.00284×T - 0.001167

0.00284×T = 0.001167

T = 0.001167/0.00284 ≈ 0.411 K

Which is 0.411  -273.15 ≈ -272.74 °C

The absolute zero temperature is -272.74 °C

2) The given volume of the gas = 20.00 dm³ at 0°C and 1.000 atm

The slope of the volume temperature graph at constant pressure = 0.0741 dm³/(°C)

The temperature is converted to Kelvin temperature, in order to apply Charles law as follows;

0°C = 0 + 273.15 K = 273.15 K

Therefore, the equation of the graph can be presented as follows;

v - 20.00 = 0.0741 × (T - 273.15)

Which gives;

v = 0.0741·T - 0.0741 ×(273.15) + 20

v = 0.0741·T - 20.2404125 + 20

v = 0.0741·T - 0.240415

Therefore at absolute 0, v = 0, we have;

0 = 0.0741·T - 0.240415

0.0741·T = 0.240415

T = 0.240415/0.0741 = 3.2445 K

The temperature in degrees is therefore;

3.2445 K - 273.15 ≈ -269.91°C

The absolute zero temperature is therefore, -269.91°C

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