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3 years ago

What process is the foundation of almost all food webs

Chemistry

1 answer:

Sergeeva-Olga [200]3 years ago

3 0

Answer:

The Sun is the major source of energy for organisms and the ecosystems of which they are a part. Producers such as plants, algae, and cyanobacteria use the energy from sunlight to make organic matter from carbon dioxide and water. This establishes the beginning of energy flow through almost all food webs

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Which is the best way to turn helium gas into liquid helium?

erma4kov [3.2K]

To create the liquid and superfluid states you cool down helium gas to a few degrees above absolute zero

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3 years ago

How does the use of hydroelectric energy compare to the use of coal?

lora16 [44]

Hydroelectric energy uses the movement of water to spin a turbine and produce electricity. Coal is a fossil fuel that is burned as an energy source, which results in emissions such as carbon dioxide, nitrogen, and sulfur. Coal use is associated with waste products both from the mining process to acquire the coal and also from the actual use of coal itself. Hydroelectric energy does not involve waste, but does pose potential problems to waterways. Coal use is more common than the use of hydroelectric energy.

4 0

3 years ago

Consider the reaction

SOVA2 [1]

Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Part (a) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}$

and,

$\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

Part (b) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}$

and,

$-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

5 0

3 years ago

The half-life of cesium-137 is 30 years. suppose we have a 200-mg sample.

Assoli18 [71]

a) To find the mass after t years:

we will use this formula:

A = Ao / 2^n

when A =the amount remaining

and Ao = the initial amount

and n = t / t(1/2)

by substitution:

∴ A = 200 mg/ 2^(t/30y)

b) Mass after 90 y :

by using the previous formula and substitute t by 90 y

A = 200mg/ 2^(90y/30y)

∴ A = 25 mg

C) Time for 1 mg remaining:

when A= Ao/ 2^(t/t(1/2)

so, by substitution:

1 mg = 200 mg / 2^(t/30y)

∴2^(t/30y) = 200 mg by solving for t

∴ t = 229 y

7 0

3 years ago

When you rub two sticks together, they get hot.

worty [1.4K]

You convert kinetic energy into thermal energy when you rub two sticks together.

6 0

3 years ago