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svetlana [45]
3 years ago
11

The Great Burdock plant’s seeds have spines on them that attach to the fur of animals that brush against it. The seed then trave

ls with the animal until it eventually falls off, which spreads the plant’s seeds farther than the plant could have done. What type of symbiotic relationship is this?
a.
Mutualism
b.
Commensalism
c.
Parasitism
d.
Competition
Chemistry
2 answers:
Sphinxa [80]3 years ago
3 0
The answer is B. Commensalism.

<span>Commensalism is a relationship between two organisms in which only one of them has benefit, and the other one is not affected. In this example, the Great Burdock's plants spread their seeds using animals, so they benefit from this relationship. On the other hand, animals neither have benefits not are harmed from the relationship.</span>
Leokris [45]3 years ago
3 0

Answer: b. Commensalism

Explanation:

Symbiotic relationship can be define as the association of two organisms belonging to different species. The two organisms either receive benefit, harm or not affected at all.

Commensalism is a type of symbiotic relationship in which the two organisms live in association. In this interaction one recieves the benefit and other one is not influeneced.

According to given example, great burdock plant is receiving benefit as it's seeds are dispersed which will cause an increase in the population of burdock. But animal which is facilitating the dispersal is unaffected.

Hence, this interaction is an example of commensalism.

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What is the period number in which helium is found
julsineya [31]

Answer:

Helium is the second element on the periodic table. It is located in period 1 and group 18 or 8A on the righthand side of the table. This group contains the noble gases, which are the most chemically inert elements on the periodic table. Each He atom has two protons and usually two neutrons and two electrons.

Explanation:

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6 0
2 years ago
Read 2 more answers
Consider the equilibrium reaction and its equilibrium constant expression. Br 2 ( g ) + 2 NO ( g ) − ⇀ ↽ − 2 NOBr ( g ) K = [ NO
zavuch27 [327]

Answer:

K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}

Explanation:

Hello,

In this case, for the equilibrium condition, the equilibrium constant is defined via the law of mass action, which states that the division between the concentrations of the products over the concentration of the reactants at equilibrium equals the equilibrium constant, for the given reaction:

2 Br_2 ( g ) + 4 NO ( g ) \rightleftharpoons  4NOBr ( g )

The suitable equilibrium constant turns out:

K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}

Or in terms of the initial equilibrium constant:

K_2=K_1^2

Since the second reaction is a doubled version of the first one.

Best regards.

5 0
3 years ago
. In which reaction is nitric acid acting as an oxidising agent?
Talja [164]

Answer:

B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2

Explanation:

Hello,

In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.

Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.

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8 0
2 years ago
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What does scientist use to deign their experiments
joja [24]

Answer:

a chart

Explanation:

a chart a chart is the answer

8 0
2 years ago
Determine the specific heat ofmaterial if a 12g sample absorbed 48j as it was heated from 20-40
devlian [24]

Answer:

c =0.2 J/g.°C

Explanation:

Given data:

Specific heat of material = ?

Mass of sample = 12 g

Heat absorbed = 48 J

Initial temperature = 20°C

Final temperature = 40°C

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT =  40°C -20°C

ΔT =  20°C

48 J = 12 g×c×20°C

48 J =240 g.°C×c

c = 48 J/240 g.°C

c =0.2 J/g.°C

6 0
2 years ago
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