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3 years ago

If there were no decomposers, which of the following would most likely happen?

Chemistry

1 answer:

alukav5142 [94]3 years ago

8 0

Answer:

B. Some plants might die due to a shortage of important nutrients.

Explanation:

Decomposers are organisms that consume dead, decaying material in the ecosystem. They can then recycle these nutrients back into the soil to be used by plants, continuing the nutrient cycle.

The amount of dead plant and animal matter would <em>increase </em>because there is nothing to breakdown the matter. Animals cannot photosynthesise under any circumstances. The lifespan of plants and animals would not be affected.

However, without decomposers recycling essential nutrients such as nitrogen back into the soil, the plants might die because they cannot perform important tasks like protein synthesis.

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Which statement best describes the polarity of the molecule Sil4?

Nadusha1986 [10]

Answer:

I dont a picture

Explanation:

but the answer is 24

6 0

3 years ago

In atmospheric chemistry, the following chemical reaction converts SO2, the predominant oxide of sulfur that comes from combusti

Misha Larkins [42]

Answer:

Explanation:

From the given information;

The chemical reaction can be well presented as follows:

$\mathtt{SO_{2(g)} + \dfrac{1}{2}O_{2(g)} }$ ⇄ $\mathtt{3SO_{2(l)}}$

Now, K is known to be the equilibrium constant and it can be represented in terms of each constituent activity:

i.e

$K = \dfrac{a_{so_3}}{a_{so_2} a_{o_2}^{\frac{1}{2}}}$

However, since we are dealing with liquids solutions;

$K = \dfrac{1}{\dfrac{Pso_2}{P^0}\Big ( \dfrac{Po_2}{P^0} \Big)^{1/2}}$ since the activity of $a_{so_3}$ is equivalent to 1

Hence, under standard conditions(i.e at a pressure of 1 bar)

$K = \dfrac{1}{Pso_2Po_2^{1/2}}$

(b)

From the CRC Handbook, we are meant to determine the value of the Gibb free energy by applying the formula:

$\Delta _{rxn} G^o = \sum \Delta_f \ G^o (products) - \sum \Delta_fG^o (reactants) \\ \\ = (1) (-368 \ kJ/mol) - (\dfrac{1}{2}) (0) - ((1) (-300.13 \ kJ/mol)) \\ \\ = -368 \ kJ/mol + 300.13 \ kJ/mol \\ \\ \simeq -68 \ kJ/mol$

Thus, for this reaction; the Gibbs frree energy = -68 kJ/mol

(c)

Le's recall that:

At equilibrium, the instantaneous free energy is usually zero &

Q(reaction quotient) is equivalent to K(equilibrium constant)

So;

$\mathtt{\Delta _{rxn} G = \Delta _{rxn} G^o + RT In Q}$

$\mathtt{0- \Delta _{rxn} G^o = RTIn K } \\ \\ \mathtt{ \Delta _{rxn} G^o = -RTIn K } \\ \\ K = e^{\dfrac{\Delta_{rxn} G^o}{RT}} \\ \\ K = e^{^{\dfrac{67900 \ J/mol}{8.314 \ J/mol \times 298 \ K}} }$

$K =7.98390356\times 10^{11} \\ \\ \mathbf{K = 7.98 \times 10^{11}}$

(d)

The direction by which the reaction will proceed can be determined if we can know the value of Q(reaction quotient).

This is because;

If Q < K, then the reaction will proceed in the right direction towards the products.

However, if Q > K , then the reaction goes to the left direction. i.e to the reactants.

So;

$Q= \dfrac{1}{Pso_2Po_2^{1/2}}$

Since we are dealing with liquids;

$Q= \dfrac{1}{1 \times 1^{1/2}}$

Q = 1

Since Q < K; Then, the reaction proceeds in the right direction.

Hence, SO2 as well O2 will combine to yield SO3, then condensation will take place to form liquid.

8 0

3 years ago

What will the pressure be if 89.9 moles of argon are contained in a 12.0 L cylinder that is pressurized at a temperature of 300

irinina [24]

P=nRT/V
p=89.9(8.314)(12)/300
P=8969.14/300
P=29.89atm
P=29.9atm

Done!

7 0

2 years ago

Which location represents a period on a Periodic Table? A. a vertical row B. a horizontal row C. a diagonal row on the per

prohojiy [21]

The periods inside the Periodic Table is B. a horizontal row, and is numbered 1 through 7

hope this helps

7 0

4 years ago

Why do scientists use spiral tubes instead of straight ones?

elixir [45]

Answer:

Heat and mass transfer of a LiBr/water absorption heat pump system (AHP) was experimentally studied during working a heating-up mode. The examination was performed for a single spiral tube, which was simulated for heat transfer tubes in an absorber. The inside and outside of the tube were subjected to a film flow of the absorption liquid and exposed to the atmosphere, respectively. The maximum temperature of the absorption liquid was observed not at the entrance but in the region a little downward from the entrance in the tube. The steam absorption rate and/or heat generation rate in the liquid film are not constant along the tube. Hence the average convective heat transfer coefficient between the liquid film flowing down and the inside wall of the tube was determined based on a logarithmic mean temperature difference between the tube surface temperature and the film temperature at the maximum temperature location and the bottom. The film heat and mass transfer coefficients rose with increasing Reynolds number of the liquid film stream.

4 0

3 years ago