A fruit fly is considered a consumer rather than a producer why

1 answer:

5 0

In an ecosystem, the only true producers are autotrophic organisms like plants and bacteria. These organisms produce energy by converting the energy from the sun into simple sugars.
All of the other organisms in the food chain, like the fruit fly, are simply consuming the energy produced by the plants/bacteria and not actually making/producing energy from a new source.

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according to the kinetic molecular theory the pressure exerted by a sample of gas at constats volume doubles when the absolute t

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Answer: gas molecules will hit the container walls more frequently and with greater force

Explanation:

According to the postulates of kinetic molecular theory:

1. The pressure exerted by a gas in a container results from collisions between the gas molecules and the container walls.

2. The average kinetic energy of the gas molecules is proportional to the kelvin temperature of the gas.

When the temperature is increased, so the average kinetic energy and the rms speed also increase. This means that the gas molecules will hit the container walls more frequently and with greater force because they are all moving faster. This increase the pressure.

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The combustion of acetylene gas is represented by this equation: 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)

MAVERICK [17]

Answer:

Approximately $2.46\; \rm mol$ .

Explanation:

Make use of the molar mass data ( $M({\rm C_2H_2}) = 26.04\; \rm g \cdot mol^{-1}$ ) to calculate the number of moles of molecules in that $64.0\; \rm g$ of $\rm C_2H_2$ :

$\begin{aligned}n({\rm C_2H_2}) &= \frac{m({\rm C_2H_2})}{M} \\ &= \frac{64.0\; \rm g}{26.04\; \rm g\cdot mol^{-1}}\approx 2.46\; \rm mol\end{aligned}$ .

Make sure that the equation for this reaction is balanced.

Coefficient of $\rm C_2H_2$ in this equation: $2$ .

Coefficient of $\rm H_2O$ in this equation: $2$ .

In other words, for every two moles of $\rm C_2H_2$ that this reaction consumes, two moles of $\rm H_2O$ would be produced.

Equivalently, for every mole of $\rm C_2H_2$ that this reaction consumes, one mole of $\rm H_2O$ would be produced.

Hence the ratio: $\displaystyle \frac{n({\rm H_2O})}{n({\rm C_2H_2})} = \frac{2}{2} = 1$ .

Apply this ratio to find the number of moles of $\rm H_2O$ that this reaction would have produced:

$\begin{aligned}n({\rm H_2O}) &= n({\rm C_2H_2}) \cdot \frac{n({\rm H_2O})}{n({\rm C_2H_2})} \\ &\approx 2.46\; \rm mol \times 1 = 2.46\; \rm mol\end{aligned}$ .