What is the name of the average weather in an area over a long period of time?

The materials created from a chemical reaction are called

sweet [91]

The answer would be products

8 0

3 years ago

Ecologists use food web diagrams, like the one shown, as tools in their studies of different populations of organisms within an

Mice21 [21]

Ecologists use food web diagrams as tools in their studies of different populations of organisms within an ecosystem because it shows the feeding relationship between organisms in the ecosystem.

This is defined as the graphical representation of the feeding relationship which occurs between organisms in the ecosystem and helps to predict the outcome in the addition or removal of a species.

The food web also helps to determine the various ways in which energy flows and it usually starts from the producers which are the plants as a result of their photosynthetic ability which involves the conversion of solar energy to chemical energy.

There are various organisms in the ecosystem and the food web helps to provide a better understanding of their feeding habits and impact various factors will result in.

Read more about Food web here brainly.com/question/4656320

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4 0

1 year ago

Consider the following reaction:

Kitty [74]

Answer : The value of equilibrium constant (Kc) is, 0.0154

Explanation :

The given chemical reaction is:

$SO_2Cl_2(g)\rightarrow SO_2(g)+Cl_2(g)$

Initial conc. $2.4\times 10^{-2}$ 0 0

At eqm. $(2.4\times 10^{-2}-x)$ x x

As we are given:

Concentration of $Cl_2$ at equilibrium = $1.3\times 10^{-2}M$

That means,

$x=1.3\times 10^{-2}M$

The expression for equilibrium constant is:

$K_c=\frac{[SO_2][Cl_2]}{[SO_2Cl_2]}$

Now put all the given values in this expression, we get:

$K_c=\frac{(x)\times (x)}{2.4\times 10^{-2}-x}$

$K_c=\frac{(1.3\times 10^{-2})\times (1.3\times 10^{-2})}{2.4\times 10^{-2}-1.3\times 10^{-2}}$

$K_c=0.0154$

Thus, the value of equilibrium constant (Kc) is, 0.0154

4 0

3 years ago

An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for

8090 [49]

Answer: The freezing point and boiling point of the solution are $-6.6^0C$ and $101.8^0C$ respectively.

Explanation:

Depression in freezing point:

$T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_f$ = freezing point of solution = ?

$T^o_f$ = freezing point of water = $0^0C$

$k_f$ = freezing point constant of water = $1.86^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_f=-6.6^0C$

Therefore,the freezing point of the solution is $-6.6^0C$

Elevation in boiling point :

$T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of water = $100^0C$

$k_b$ = boiling point constant of water = $0.52^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_b=101.8^0C$

Thus the boiling point of the solution is $101.8^0C$

4 0

3 years ago

Good Morning I have a question I need help and can not find the answer o it maybe someone help me? The question is _______ Most

Semenov [28]

Answer:

hydro, water

Explanation:

3 0

3 years ago