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Mama L [17]
3 years ago
10

Why a slow growing forest can have a very low NPP and yet store a massive amount of biomass.

Chemistry
1 answer:
drek231 [11]3 years ago
8 0
Net Primary Productivity ... the amount of biomass present in an ecosystem at a particular time .... Explain why a slow growing forest can have a very low NPP and yet store a massive amount of biomass.
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Really strong acids and bases are found at the​
makkiz [27]

Ends....................?

4 0
3 years ago
What element has the electron configuration 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p2?
Marina CMI [18]
<span>6s²4f¹⁴5d¹⁰6p²
6 shows that the element is in the 6 period,
6p² shows that the element is in the 14th group. (1 and 2 groups have s -electrons as last ones, 13 group has s²p¹, and 14 group has s²p²)
The element is Pb.

</span>
3 0
3 years ago
Read 2 more answers
The molar solubility of silver bromide, AgBr in pure water is 0.0007350 mol/L. What is the
gayaneshka [121]

Answer:

0.000000540

Explanation:

Step 1: Make an ICE chart for the solution of AgBr

"S" represents the molar solubility of AgBr

        AgBr(s) ⇄ Ag⁺(aq) + Br⁻(aq)

I                           0             0

C                          +S          +S

E                           S             S

Step 2: Write the expression for the solubility product constant (Ksp)

Ksp = [Ag⁺] [Br⁻] = S × S

Ksp = S² = (0.0007350)² = 0.000000540

7 0
3 years ago
The question is in the picture below
Rus_ich [418]

Answer:

\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3

Explanation:

Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.

With Hess's Law in mind, let's see how A can be converted to 2C +E.

\bf{\text{A} \rightarrow 2\text{B}}                  (Δ\text{H}_1)  -----(1)

Since we have 2B, multiply the whole of II. by 2:

\bf{2\text{B} \rightarrow 2\text{C} +2\text{D}}       (2Δ\text{H}_2) -----(2)

This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is \text{A} \rightarrow 2\text{C} +2\text{D}.

Reversing III. gives us a negative enthalpy change as such:

\bf{2\text{D} \rightarrow \text{E}}                  (-Δ\text{H}_3) -----(3)

This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of \text{A} \rightarrow 2\text{C} +\text{E}, which is also the equation of interest.

Adding all three together:

\text{A} \rightarrow 2\text{C}+\text{E}            (\bf{\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3 })

Thus, the first option is the correct answer.

Supplementary:

To learn more about Hess's Law, do check out: brainly.com/question/26491956

4 0
1 year ago
Which situation is the best example of interia
AveGali [126]
One's body movement to the side when a car makes a sharp turn. Tightening of seat belts in a car when it stops quickly. A ball rolling down a hill will continue to roll unless friction or another force stops it.
6 0
3 years ago
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