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3 years ago

What is the 10% rule? What is its significance? Why is energy lost?

1 answer:

4 0

The 10% rule means that approximately there will be only 10% energy transferred from a trophic level into another. That was because the consumer on the upper level will use the energy before it was consumed. About 90% of the energy used so only 10% remain will be transferred.
Example: an antler eating grass with a total 100 calories. Then the antler is moving and growing, using 90 calories. Then a lion eats the antler, but the energy remains is only 10 calorie because 90 calories are already used.

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Considering the combustion reaction of propane again, what mass of propane (C3Hg) is necessary to react with 4.53 g of oxygen? C

zimovet [89]

Answer:

1.25 g C3H8

Explanation:

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7 0

2 years ago

A 1.1000 gram carbonate sample chosen from Li2CO3, K2CO3, Na2CO3 and CaCO3 was reacted with H2SO4 and was found to lose 0.3497 g

Stels [109]

Answer:

For 1: The percentage composition of $CO_2$ in unknown carbonate sample is 31.79 %

For 2: The unknown carbonate sample is potassium carbonate.

Explanation:

To calculate the percentage composition of carbon dioxide in sample, we use the equation:

$\%\text{ composition of }CO_2=\frac{\text{Mass of }CO_2}{\text{Mass of sample}}\times 100$ .....(1)

For 1:

Mass of sample = 1.1000 g

Mass of carbon dioxide = 0.3497 g

Putting values in equation 1, we get:

$\%\text{ composition of }CO_2=\frac{0.3497g}{1.1000g}\times 100=31.79\%$

Hence, the percentage composition of $CO_2$ in unknown carbonate sample is 31.79 %

For 2:

We are given:

Mass of carbon dioxide = $[1\times 12.01)+(2\times 16.00)]=44.01g$

For lithium carbonate:

Mass of lithium carbonate= 73.892 g

Mass of carbon dioxide = 44.01 g

Putting values in equation 1, we get:

$\%\text{ composition of }CO_2=\frac{44.01g}{73.892g}\times 100=59.55\%$

The percent composition of carbon dioxide in lithium carbonate is 59.55 %

For potassium carbonate:

Mass of potassium carbonate= 138.21 g

Mass of carbon dioxide = 44.01 g

Putting values in equation 1, we get:

$\%\text{ composition of }CO_2=\frac{44.01g}{138.21g}\times 100=31.84\%$

The percent composition of carbon dioxide in potassium carbonate is 31.84 %

For sodium carbonate:

Mass of sodium carbonate= 105.99 g

Mass of carbon dioxide = 44.01 g

Putting values in equation 1, we get:

$\%\text{ composition of }CO_2=\frac{44.01g}{105.99g}\times 100=41.52\%$

The percent composition of carbon dioxide in sodium carbonate is 41.52 %

For calcium carbonate:

Mass of calcium carbonate = 100.09 g

Mass of carbon dioxide = 44.01 g

Putting values in equation 1, we get:

$\%\text{ composition of }CO_2=\frac{44.01g}{100.09g}\times 100=43.97\%$

The percent composition of carbon dioxide in calcium carbonate is 43.97 %

As, the percent composition of carbon dioxide in the unknown sample matches the percent composition of carbon dioxide in potassium carbonate.

Hence, the unknown carbonate sample is potassium carbonate.

7 0

3 years ago

If we dissolve 27 g of salt in 249 g of water, what is the mass of the resulting solution?

zavuch27 [327]

Answer:

276g

Explanation:you are adding salt to water the mass is not destroyed so you just add the numbers hope this helps god bless

7 0

3 years ago

How many moles are in 325mL of 1.25M (NH4)2SO4

mote1985 [20]

Answer:

$n_{solute}=0.406mol(NH_4)_2SO_4$

Explanation:

Hello,

In this case, since the molarity is defined as the ratio of the moles of the solute to the volume of the solution in liters:

$M=\frac{n_{solute}}{V_{solution}}$

For the given volume and molarity, we solve for the moles of ammonium sulfate (solute) as shown below:

$n_{solute}=M*V_{solution}=1.25\frac{mol}{L}*325mL*\frac{1L}{1000mL} \\\\n_{solute}=0.406mol(NH_4)_2SO_4$

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4 0

3 years ago

Over time the Mexican-Free Tailed bat has evolved to be able to echolocate in the dark to find prey. Bats are able to see, but d

Law Incorporation [45]

Answer:

Light

Explanation:

1 They were not albe to see at night because it was dark so the lack of light caused them to use echolocation.

Hope this helps

4 0

3 years ago