All categories

3 years ago

How does the lithosphere affect wildfires

1 answer:

6 0

Sphere to Sphere Interactions
The fire burns the trees(Biosphere) which leaves the ash which becomes part of the soil (Lithosphere) where new growth(Biosphere) can happen.

You might be interested in

On a mission to a newly discovered planet, an astronaut finds gallium abundances of 61.29% for 69Ga and 39.71% for 71Ga. What is

Verizon [17]

(exact weight of isotope #1) (abundance of isotope #1) + (exact weight of isotope #2) (abundance of isotope #2) = average atomic weight of the element(68.72*61.29) + (39.71*70.92) = 4211.84 + 2816.237028.07/100= 70.28

70.28 is the atomic mass of Gallium for the location

4 0

3 years ago

explain what happens if high pressure is implemented on the equilibrium reaction . State your observations for both rate of reac

amm1812

The equilibrium position will shift in order to counterbalance the change. That means the equilibrium position will shift, lowering the pressure once more.... When the pressure on a gas reaction is increased, the equilibrium moves to the side with fewer molecules.

8 0

2 years ago

Consider the following reactions: 1. 2 SO3(g) ⇄ 2 SO2(g) + O2(g) K 1 = 2.3 × 10-7 2. 2 NO3(g) ⇄ 2 NO2(g) + O2 (g) K 2 = 1.4× 10-

juin [17]

Answer:

K =78

Explanation:

Step 1: Data given

2SO3(g) <--> 2SO2(g) + O2(g) kc = 2.3 x 10^-7

2NO3(g) <--> 2NO2(g) + )2(g) kc = 1.4 x 10^-3

Step 2: Calculate K

Lets write out the two reactions in the proper order and look at how they sum together:

2 SO2(g) + O2(g) <---> 2 SO3(g) (1)

2NO3(g) <---> 2 NO2(g) + O2(g) (2)

The two reactions as now written give us the correct reactants and products on the correct sides of the reaction arrow.

Since we have O2 as both a reactant and a product, we can cancel O2 and are not part of the final overall reaction equation.

Koverall = Kc1 * Kc2

Because we reversed reaction number 1 this affects its Kc via the following:

Krev = 1/Kfwd.

We then replace Kc1 with its value for the reverse direction.

So Koverall now = (1/Kfwd) * Kc2

The sum of the two reactions above gives us:

2 SO2(g) + 2 NO3(g) <---> 2 SO3(g) + 2 NO(g)

The problem states to give the K value for the reaction where all the numbers in front of the molecules are (1), and we have (2)'s. So basically if we multiply the whole reaction by 1/2 we'll get the final overall equation we want.

1/2 ( 2 SO2(g) + 2 NO3(g) <----> 2 SO3(g) + 2 NO(g) )

So Kfinal = (Koverall)^1/2

K = ( 1/Kfwd * Kc2)^1/2

K = ( [1 / 2.3 * 10^-7] * 1.4 * 10^-3)^1/2

K =78

3 0

3 years ago

Based on Figure 2, which statement best describes what happens to the amount of matter in the reaction?

scoray [572]

Answer:15 +18

Explanation:15 fuel +18 oxygen + 33 cd

8 0

2 years ago

A 1.00 g sample of a hydrogen peroxide (H2O2) solution is placed in an Erlenmeyer flask and diluted with 20 mL of 1 M aqueous su

Minchanka [31]

Following reaction is involved in present system:

2KMnO4 + 5H2O2 + 3H2SO4 → 2MnSO4 + K2SO4 + 5O2 + 8H2O

From the above balance reaction, it can be seen that 2 moles of KMnO4 is consumed for every 5 moles of H2O2.

Now, percent by mass of hydrogen peroxide in the original solution can be estimated as follows:
percent by mass = $\frac{\text{mass of H2O2(g)}}{\text(volume of H2SO4(ml))}X 100$
∴percent by mass = $\frac{\text{1}}{\text(25)}X 100$
= 4 %

5 0

3 years ago