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navik [9.2K]
3 years ago
12

Which is an example of an interaction between the biosphere and the atmosphere?

Chemistry
1 answer:
son4ous [18]3 years ago
6 0
Desalination morning
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At –45oC, 71 g of fluorine gas take up 6843 mL of space. What is the pressure of the gas, in kPa?
Anika [276]
The moles of fluorine present are 71/19 = 3.74
Now, we know that one mole of gas at 273 K and 101.3 kPa (S.T.P.) occupies 22.4 liters
Volume of 3.74 moles at S.T.P = 3.74 x 22.4
Volume = 83.776 L = 83,776 mL

Now, we use Boyle's law, that for a given amount of gas,
PV = constant

P x 6843 = 101.3 x 83776
P = 1,240 kPa
4 0
3 years ago
How do you draw a bohr diagram for a boron ion?
Otrada [13]
Because Boron likes to lose 3 electrons when it undergoes ionization, we draw a boron ion like a helium atom, with just 2 electrons in the first shell, and 0 in the second

5 0
2 years ago
Read 2 more answers
What does geocentric mean
motikmotik

Answer:

Relating to, measured from, or as if observed from the earth's center.

5 0
2 years ago
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Potassium permanganate as a symbol
Alexxx [7]

Answer: KMnO4-

Explanation:

You're looking at one potassium plus a polyatomic ion.

So K plus MnO4, equals:

KMnO4-

It also has a molar mass of 158.04 g/mol, I don't know if you need that, but I thought it would be nice to include it.

7 0
2 years ago
At 700 K, the reaction2SO2(g) + O2(g) 2SO3(g)has the equilibrium constant Kc = 4.3 x 106, and the following concentrations are p
dedylja [7]

Explanation:

The given reaction is as follows.

       2SO_{2} + O_{2}(g) \rightarrow 2SO_{3}(g)

Value of equilibrium constant is given as K_{c} = 4.3 \times 10^{6}[/tex].

Concentration of given species is [SO_2] = 0.010 M; [SO_3] = 10.M; [O_2] = 0.010 M.

Formula for experimental value of equilibrium constant (Q) is as follows.

             Q = \frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}

Putting the given concentration as follows.

              Q = \frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}

             Q = \frac{(10)^{2}}{(0.010)^{2}(0.010)}

              Q = 10^{8}

It is known that when Q > K_{eq}, then reaction moves in the backward direction.

When Q < K_{eq}, then reaction moves in the forward direction.

When Q = K_{eq}, then reaction is at equilibrium.

As, for the given reaction Q > K_{eq} then it means reaction moves in the backward direction.

Thus, we can conclude that the reaction is moving in the backward direction, that is, right to left to reach the equilibrium.

5 0
3 years ago
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