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3 years ago

7

Give two examples. How does predation affect an ecosystem?

2 answers:

Gnom [1K]3 years ago

8 0

Answer:

It affects the amount of a species that may be hunted by said predator, controlled you can say.

It can cause evolution in prey that causes them to not be caught, Darwins theory in a way.

Explanation:

nevsk [136]3 years ago

6 0

Explanation:

At the level of the community, predation reduces the number of individuals in the prey population. The best example of predation involve carnivorous interactions, in which one animal consumes another, Think of wolves hunting moose, owls hunting mice, or shrews hunting worms and insects.

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sertanlavr [38]

The answers are :
1 - F
2- T

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2 years ago

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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.

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3 years ago

What does kcal mean again?

Neporo4naja [7]

It means kilocalorie

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3 years ago

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Complete and balance the following redox reaction in basic solution

inn [45]

Answer:

balanced in ACID not BASE

Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O

Answer

Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O

Explanation:

Cr2O7^2-(aq) + Hg(l) ----> Hg^2+(aqH) + Cr^3+(aq)

add H^1+ (acid) to capture the O and make 7 water molecules

Cr2O7^2-(aq) + Hg(l) + H^1+ ----> Hg^2+(aqH) + Cr^3+(aq) + 7H2O

Cr goes from +6 to +3 by gaining 3 e

Hg goes from 0 to +2 by losing 2 e

we need 3 Hg for every 2 Cr

so

Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O

2 Cr on the right and left

Net 12 positive charges on the right and the left

3 Hg on the right and left

14 H on the right and left

the equation is balanced

we cannot balance the equation in a basic solution with OH^1-

we have plenty of O in the dichromate ion. we need to convert it to water which take free H^1+ from the acid

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2 years ago

If you have 85.0 grams of CO₂, calculate how many moles of CO₂ would you have?

dedylja [7]

N=m(g)/m.wt
n=85/12(1)+16(2) =1.93 moles

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2 years ago

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