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3 years ago

1. What is the relationship between ecological succession and equlibrium in an ecosystem?

Chemistry

1 answer:

alexandr1967 [171]3 years ago

4 0

Answer:

Equilibrium in ecology refers to a state that occurs such that a small disturbance or change is counter balanced by another change so that the community is restored to its original state. Thus, as a community goes through multiple changes through each stage of succession, it is not in equilibrium.

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A student dissolves of glucose in of a solvent with a density of . The student notices that the volume of the solvent does not c

nikitadnepr [17]

Answer:

0.052 M

0.059 m

Explanation:

There is some missing info. I think this is the complete question.

<em>A student dissolves 4.6 g of glucose in 500 mL of a solvent with a density of 0.87 g/mL. The student notices that the volume of the solvent does not change when the glucose dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to 2 significant digits.</em>

Step 1: Calculate the moles of glucose (solute)

The molar mass of glucose is 180.16 g/mol.

4.6 g × 1 mol/180.16 g = 0.026 mol

Step 2: Calculate the molarity of the solution

0.026 moles of glucose are dissolved in 500 mL (0.500 L) of solution. We will use the definition of molarity.

M = moles of solute / liters of solution

M = 0.026 mol / 0.500 L = 0.052 M

Step 3: Calculate the mass corresponding to 500 mL of the solvent

The solvent has a density of 0.87 g/mL.

500 mL × 0.87 g/mL = 435 g = 0.44 kg

Step 4: Calculate the molality of the solution

We will use the definition of molality.

m = moles of solute / kilograms of solvent

m = 0.026 mol / 0.44 kg = 0.059 m

4 0

3 years ago

A spectator ion is (Select all that apply.) a. an ionic component of a reactant that is unchanged by the reaction your eye, b. c

DENIUS [597]

Correct Question:

A spectator ion is (Select all that apply.)

- a piece of french fry contaminating the reaction mixture

- an ionic component of a reactant that is unchanged by the reaction

-in this experiment, nitrate ion

- your eye, carefully watching the progress of the reaction

Answer:

- an ionic component of a reactant that is unchanged by the reaction

Explanation:

A spectator ion is an ion that exists as a reactant and a product in a chemical equation. A spectator ion is one that exists in the same form on both the reactant and product sides of a chemical reaction.

Spectator ions are ions that are present in a solution but don't take part in the reaction. When reactants dissociate into ions, some of the ions may combine to form a new compound. The other ions don't take part in this chemical reaction and are therefore called spectator ions.

The correct option is therefore the option;

- an ionic component of a reactant that is unchanged by the reaction

5 0

3 years ago

We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

NeX [460]

Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

5 0

3 years ago

A block of chocolate is melting on a hot pan. What type of heat transfer is this? Describe the way heat is transferred between t

klio [65]

Answer:liquid= _________ than the next ________ is heat

Explanation:

8 0

3 years ago

In acidic solution, the sulfate ion can be used to react with a number of metal ions. One such reaction is SO42−(aq)+Sn2+(aq)→H2

allsm [11]

Answer:

The final balanced equation is :

$SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)$

Explanation:

$SO_4^{2-}(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+Sn^{4+}(aq)$

Balancing in acidic medium:

First we will determine the oxidation and reduction reaction from the givne reaction :

Oxidation:

$Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)$

Balance the charge by adding 2 electrons on product side:

$Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)+2e^-$ ....[1]

Reduction :

$SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)$

Balance O by adding water on required side:

$SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)$

Now, balance H by adding $H^+$ on the required side:

$SO_4^{2-}(aq)+4H^+(aq)\rightarrow H_2SO_3(aq)+H_2O(l)$

At last balance the charge by adding electrons on the side where positive charge is more:

$SO_4^{2-}(aq)+4H^+(aq)+2e^-\rightarrow H_2SO_3(aq)+H_2O(l)$ ..[2]

Adding [1] and [2]:

$SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)$

The final balanced equation is :

$SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)$

4 0

3 years ago