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3 years ago

How do organisms come to live o newly formed volcanic islands

1 answer:

6 0

Answer is: small animals, insects and seeds originating form nearby land masses come to new volcanic island blown over by winds, crashing on shore pushed by the waves. Birds come easy by flying. They adopt to new enviroment or evolve into the new species. This new species are unique.

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Which of these would be found in a wetland?

andrezito [222]

Answer:... in this set (11). Which type of wetland has large numbers of cattails and grasslike plants? ... Swamp. Mangrove trees would most likely be found in which wetland? ... spring rains. Sometimes called the "River of Grass," ---- is a vast wetland.

Explanation:

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4 0

3 years ago

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Calculate the masses of oxygen and nitrogen that are dissolved in of aqueous solution in equilibrium with air at 25 °C and 760 T

shtirl [24]

Explanation:

Let us assume that the volume of given aqueous solution is 7.5 L.

Therefore, according to Henry's law, the relation between concentration and pressure is as follows.

C = $\frac{P}{K_{h}}$

where, pressure (P) = 760 torr = 1 atm

According to Henry's law, constants for gases in water at $25^{o}C$ are as follows.

$p(O_{2})$ = 0.21 atm = 0.21 bar

$p(N_{2})$ = 0.78 atm = 0.78 bar

$K_{h}$ for $O_{2}$ = $7.9 \times 10^{2} bar/mol$

$K_{h}$ for $N_{2}$ = $1.6 \times 10^{3} bar /mol$

Since, 21% oxygen is present in air so, its mass will be 0.21 g. Similarly, 78% nitrogen means the mass of nitrogen is 0.78 g.

Therefore, concebtrations will be calculated as follows.

$C(O_{2}) = \frac{0.21}{7.9 \times 10^{2}} = 2.66 \times 10^-4 mol/L$

$C(N_2) = \frac{0.78}{1.6 \times 10^3} = 4.875 \times 10^-4 mol/L$

Now, we will calculate the number of moles as follows.

$n(O_{2}) = 7.5 \times 2.66 \times 10^-4 = 1.995 \times 10^-3$ mol

$n(N_2) = 7.5 \times 4.875 \times 10^-4 = 3.66 \times 10^-3$ mol

As the molar mass of $O_2$ = 32 g/mol

Hence, mass of oxygen will be as follows.

Mass of $O_2 = 32 \times 1.995 \times 10^-3 \times 1000$

= 63.84 mg

As the molar mass of $N_{2}$ = 28

Mass of $N_{2} = 28 \times 3.66 \times 10^-3$ = 102.5 mg

Thus, we can conclude that mass of oxygen is 63.84 mg and nitrogen is 102.5 mg.

5 0

4 years ago

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The solubility of KClO3(s) in water increases as the

algol [13]

Answer: option (1) solubility of the solution increases.


Justification:


The solubility of substances in a given solvent is temperature dependent.


The most common behavior of the solubility of salts in water is that the solubiilty increases as the temperature increase.


To predict with certainty the solubility at different temperatures you need the product solubility constants (Kps), which is a constant of equlibrium of the dissolution of a ionic compound slightly soluble in water, or a chart (usually experimental chart) showing the solubilities at different temperatures.


KClO₃ is a highly soluble in water, so you do not work with Kps.


You need the solubility chart or just assume that it has the normal behavior of the most common salts. You might know from ordinary experience that you can dissolve more sodium chloride (table salt) in water when the water is hot. That is the same with KClO₃.

The solubility chart of KlO₃ is almost a straight line (slightly curved upward), with positive slope (ascending from left to right) meaning that the higher the temperature the more the amount of salt that can be dissolved.

4 0

4 years ago

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The goal of this lesson:

mote1985 [20]

Answer:

1. The concentration of N₂O₄ decrease as the temperature of the system increased.

2. The formation of products was favored by the addition of heat.

3. The reaction going from right to left is exothermic.

4. N₂O₄ → 2NO₂; HR = +14 kcal.

Explanation:

Equation of the reaction is as follows: N₂O₄ ⇄ 2NO₂

The forward reaction proceeds with an increase in temperature. As the temperature of the reaction was increased, more of the N₂O₄ decomposed to form NO₂. Therefore, the concentration of N₂O₄ deceased.

2. The formation of products, that is the forward reaction leading to formation of NO₂ was favored by the addition of heat. Thus, with increase in temperature, the concentration of NO₂ increased.

3. An endothermic reaction is a type of reaction which requires energy input in the form of heat in order to proceed.

From the observations during the reaction, an increase in temperature by the addition of heat resulted in an increase in the forward reaction, therefore, the forward reaction is endothermic and the backward reaction is exothermic. Thus, the reaction in which colorless N₂O₄ is produced, is an exothermic reaction.

4. The change in enthalpy of a reaction is the difference in the heat content of reactants and products. For exothermic reactions, enthalpy change is negative, whereas for endothermic reactions, enthalpy change is positive.

The decomposition of N₂O₄ to NO₂ is an endothermic reaction. Hence, the correct chemical equation is: N₂O₄ → 2NO₂; HR = +14 kcal.

7 0

3 years ago

In your own words, explain how gametes (the cells made for sexual reproduction) differ from normal body cells.

NNADVOKAT [17]

Answer:

gametes have to become haploid - they have to reduce their genetic material to a single copy, while somatic cells are diploid

Explanation:

7 0

3 years ago

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