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2 years ago

If the manta ray gets 50 kcal of energy by eating the starfish, how much energy does the clam get from eating the plankton?

Chemistry

2 answers:

taurus [48]2 years ago

8 0

Answer:

5,000 kcal

Explanation:

Just took the quiz

julia-pushkina [17]2 years ago

4 0

Answer: Clam get 5000 Kcal by eating the plankton

Explanation:

In any food chain the amount of energy transferred to the successive trophic level is ten percent of the energy available at the previous trophic level.

Here, manta ray gets 50 kcal of energy by eating the starfish

The amount of energy contained in star fish is equal to Kcal

Now Starfish gets its energy from clam.

Therefore total energy in clam is equal to

Therefore, clam get 5000 Kcal by eating the plankton

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A mixture of N2, O2, and Ar has mole fractions of 0.25, 0.65, and 0.10, respectively. What is the pressure of N2 if the total pr

s344n2d4d5 [400]

Answer:

Partial pressure of nitrogen gas is 0.98 bar.

Explanation:

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gases.

$P=p_{N_2}+p_{O_2}+p_{Ar}$

$p_{N_2}=P\times \chi_{N_2}$

$p_{O_2}=P\times \chi_{O_2}$

$p_{Ar}=P\times \chi_{Ar}$

where,

$P$ = total pressure = 3.9 bar

$p_{N_2}$ = partial pressure of nitrogen gas

$p_{O_2}$ = partial pressure of oxygen gas

$p_{Ar}$ = partial pressure of argon gases

$\chi_{N_2}$ = Mole fraction of nitrogen gas = 0.25

$\chi_{O_2}$ = Mole fraction of oxygen gas = 0.65

$\chi_{Ar}$ = Mole fraction of argon gases = 0.10

Partial pressure of nitrogen gas :

$p_{N_2}=P\times \chi_{N_2}=3.9 bar\times 0.25 =0.98 bar$

Partial pressure of oxygen gas :

$p_{N_2}=P\times \chi_{O_2}=3.9 bar\times 0.65=2.54 bar$

Partial pressure of argon gas :

$p_{N_2}=P\times \chi_{Ar}=3.9 bar\times 0.10=0.39 bar$

7 0

2 years ago

Both black and white road surfaces absorb sunlight. the warmer road surface at the end of a sunny day is the

Assoli18 [71]

The warmer road surface at the end of a sunny day is the black road because during the day it absorbed more radiation (sunlight) than the withe one

5 0

2 years ago

Lars observes a substance to be a solid and to float in water at room temperature (23°C). Based on the given properties, which s

Triss [41]

Answer:

D. Sulfur Hexafluride

Explanation:

above it says the substance floats above water at room temperature and lists some substances and their density at room temp!
we know that the density of water is 1.0 so the substance in order for it to float has to be less than 1.0 and the densities for Sulfer Hexa, are all less than 1!!

I hope this helped !!

4 0

2 years ago

What is the mole fraction of potassium dichromate, K2Cr2O7, in a solution prepared from 24.42 g of potassium dichromate and 240.

Harrizon [31]

Answer:

$6.19\times 10^{-3}$ is the mole fraction of potassium dichromate.

Explanation:

Mass of potassium dichromate = 24.42 g

Moles of potassium dichromate = $n_1=\frac{24.42 g}{294.185 g/mol}=0.0830 mol$

Mass of water = 240.0 g

Moles of water = $n_2=\frac{240.0 g}{18.015 g/mol}=13.3222 mol$

Mole fraction is calculated by:

$\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{0.0830 mol}{0.0830 mol+13.3222 mol}=0.00619=6.19\times 10^{-3}$

$6.19\times 10^{-3}$ is the mole fraction of potassium dichromate.

8 0

2 years ago

1. The specific heat capacity of iron is 0.461 J g–1 K–1 and that of titanium is 0.544 J g–1 K–1. A sample consisting of a mixtu

mezya [45]

Answer:

The answer is 80,1 °C

Explanation:

Let´s start from the mass of the sample and the heat capacities:

First of all, we must calculate an average heat capacity. That's because we have a mixture and it is unknown the heat capacity of the whole sample.

The way we should do this calculation is as follows:

(1) $H_{average}=Mass Fraction_{first component}* H_{first component}+MassFraction_{secondcomponent}*H_{second component}$

For example, the mass fraction of Fe is simply:

(2) $MassFraction_{Fe}=\frac{10g Fe}{10g Fe + 10gTi}=0.5$

If you combine the equations (1) and (2) you have:

(3) $H_{average}=0.5*0.461+0.5*0.544=0.5025\frac{J}{g-K}$

Once calculated the average heat capacity we can solve the problem taking into account the corresponding equation:

(4) $Q=m*H_{average}*(T_2-T_1)$

Remember that:

Q: Heat gained or lost

m: Mass of the sample you want to analize

$H_{average}$ : The value obtained in equation (3)

$T_2$ : Final temperature of the sample

$T_1$ : Initial temperature of the sample

Now we must replace the problem data in equation (4)

Take into account:

Heat gained in a system have a positive value
Heat lost in a system have a negative value

In this problem the sample loses 200 J, for this reason $Q=-200J$

The mass of the whole sample is: 10g of Fe + 10g of Ti = 20g of sample
The temperatures must be in absolute units of temperature (these are: rankine or kelvin)
The initial temperature of the system is 100°C or 373K

Now we are ready to use equation (4):

(5) $-200J=20g*0.5025\frac{J}{g*K} *(T_2-373K)$

It is clear that the unknown in equation (5) is $T_2$

The next step is to calculate $T_2$ . Don't forget the signs; these are important.

Key concept: Since the system is loosing heat, the final temperature of the system ( $T_2$ ) should be lower than the initial temperature ( $T_1$ )

7 0

2 years ago

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