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never [62]
3 years ago
9

Can someone help me with this???​

Chemistry
1 answer:
vivado [14]3 years ago
3 0
H^5C>4 Im pretty sure is like that
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N2 (g) + 2O2(g) = 2NO2 (g) ΔH = 66.4 kJ 2NO (g) + O2 (g) = 2NO2 (g) ΔH = -114.2 kJ the enthalpy of the reaction of the nitrogen
RSB [31]

Answer:

ΔH  = 180.6 kJ

Explanation:

Given that:

N2 (g) + 2O2(g) = 2NO2 (g)           ΔH = 66.4 kJ

<u>2NO (g) + O2 (g) = 2NO2 (g)         ΔH = -114.2 kJ                     </u>

N2 (g) + O2 (g) = 2NO (g)              ΔH  = ????

The subtraction of both equations would yield the unknown ΔH , therefore:

ΔH = 66.4 - ( - 114.2 kJ)

ΔH  = 180.6 kJ

3 0
3 years ago
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Which of the following adaptations would you expect tofind in an animal in the alpine biome, but not in the taiga
almond37 [142]
I think the best answer is migration. In an alpine biome, animals have two problems which is the very cold temperature and the UV rays from the sun. Most alpine animals hibernate and migrate to warmer areas of the mountains when the temperature continues to drop.
7 0
3 years ago
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A chemistry student must write down in her lab notebook the concentration of a solution of potassium chloride. The concentration
DerKrebs [107]

Answer:

3.65 g / ml correct to 3 sig. fig.

Explanation:

The computation of the concentration required is shown below:

As we know that

[A] = mass of solute ÷ volume of solution

Before that first find the mass of solute

Given that

Initial weight = 5.55g

And,

Final weight = 92.7 g

So,

Mass of KCl is

= 92.7 - 5.55

= 87.15 g ~ 87.2 g

Now the KCi is fully dissolved, so the volume is 23.9 ml

So,  concentration is

= 87.2 g ÷ 23.9 ml

= 3.65 g / ml correct to 3 sig. fig.

6 0
3 years ago
Aluminum has a density of 2.70 g/m. Calculate the mass (in grams) of a piece of aluminum having a volume of 382 mL
Natalka [10]

The aluminum has a mass of 1031.4 grams.

8 0
3 years ago
The half-life of cobalt-60 is 5.3 years. after __________ years, 1/4 of the original amount of cobalt-60 will remain.
cestrela7 [59]
The  number of  years  required  for 1/4  cobalt-60  to remain   after   decay  is calculated  as follows

  after  one  half life  1/2 of the original  mass isotope  remains

after  another half life 1/4  mass  of  original  mass  remains

therefore  if one  half  life is  5.3   years  then  the  years required

= 2  x 5.3years =  10.6   years
4 0
3 years ago
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