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3 years ago

15

Check your understanding: Which layer of the atmosphere is capable of supporting life??? a) troposphere,,b) stratosphere,,c) mes

osphere,,d) thermosphere...which is the correct answer

1 answer:

denis-greek [22]3 years ago

4 0

Troposphere according to this image

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Plz help. I just need someone to check answer. :(

My name is Ann [436]

These substances can be separated by distillation, so your answer is A.

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3 years ago

Thermal energy lab report

Aleonysh [2.5K]

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4 years ago

Some one help me with this will give brainliest

Alona [7]

14. C
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3 years ago

Ammonia reacts with diatomic oxygen to form nitric oxide and water vapor: 4 NH3 + 5 O2 → 4 NO + 6 H2O When 40.0 g NH3 and 50.0 g

luda_lava [24]

Answer:

18.75 g of NH3.

Explanation:

The balanced equation for the reaction is given below:

4NH3 + 5O2 → 4NO + 6H2O

Next, we shall determine the masses of NH3 and O2 that reacted from the balanced equation.

This can be obtained as follow:

Molar mass of NH3 = 14 + (3x1) = 17 g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160 g

From the balanced equation above,

68 g if NH3 reacted with 160 g of O2.

Next, we shall determine the excess reactant. This can be obtained as follow:

From the balanced equation above,

68 g if NH3 reacted with 160 g of O2.

Therefore, 40 g of NH3 will react with = (40 × 160)/68 = 94.12 g of O2.

From the calculations made above, we can see that it will take a higher amount of O2 i.e 94.12g than what was given i.e 50g to react completely with 40 g of NH3.

Therefore, O2 is the limiting reactant and NH3 is the excess reactant.

Next we shall determine the mass of excess reactant that reacted. This can be obtained as follow:

From the balanced equation above,

68 g if NH3 reacted with 160 g of O2.

Therefore, Xg of NH3 will react with 50 g of O2 i.e

Xg of NH3 = (68 × 50)/160

Xg of NH3 = 21.25 g

Therefore, 21.25 g of NH3 (excess reactant) were consumed in the reaction.

Finally, we shall determine mass of the remaining excess reactant as follow:

Mass of excess reactant = 40 g

Mass of excess reactant that reacted = 21.25 g

Mass of excess reactant remainig =?

Mass of excess reactant remainig = (Mass of excess reactant) – (Mass of excess reactant that reacted)

Mass of excess reactant remainig

= 40 – 21.25

= 18.75 g

Therefore, the mass of excess reactant remaining is 18.75 g of NH3.

8 0

3 years ago

t 745 K, the reaction below has an equilibrium constant (Kc) of 5.00 × 102. H2 (g) + I2 (g) ⇌ 2 HI (g) If a mixture of 0.10 mol

spayn [35]

Answer : The concentration of HI (g) at equilibrium is, 0.643 M

Explanation :

The given chemical reaction is:

$H_2(g)+I_2(g)\rightarrow 2HI(g)$

Initial conc. 0.10 0.10 0.50

At eqm. (0.10-x) (0.10-x) (0.50+2x)

As we are given:

$K_c=5.00\times 10^2$

The expression for equilibrium constant is:

$K_c=\frac{[HI]^2}{[H_2][I_2]}$

Now put all the given values in this expression, we get:

$5.00\times 10^2=\frac{(0.50+2x)^2}{(0.10-x)\times (0.10-x)}$

x = 0.0713 and x = 0.134

We are neglecting value of x = 0.134 because the equilibrium concentration can not be more than initial concentration.

Thus, we are taking value of x = 0.0713

The concentration of HI (g) at equilibrium = (0.50+2x) = [0.50+2(0.0713)] = 0.643 M

Thus, the concentration of HI (g) at equilibrium is, 0.643 M

8 0

4 years ago