All categories

3 years ago

Where does the rest of the energy “lost” go? (In food webs)

Chemistry

2 answers:

Blizzard [7]3 years ago

3 0

Explanation:

The rest of the energy is passed on as food to the next level of the food chain. The figure at the left shows energy flow in a simple food chain. Notice that at each level of the food chain, about 90% of the energy is lost in the form of heat.

damaskus [11]3 years ago

3 0

Answer:

While the total energy of a system is always conserved, the kinetic energy carried by the moving objects is not always conserved. In an inelastic collision, energy is lost to the environment, transferred into other forms such as heat.

Explanation:

You might be interested in

How many orbitals surround the nucleus in a neutral atom of sulfur (S)?

Leto [7]

Explanation:

The atomic number of S is $16$

So,number of electrons in S is $16$

The electronic configuration of S is $2,8,6$

The orbital electronic configuration of S is $^{1}s_{2}^\text{ }^{2}s_{2}^\text{ }^{2}p_{x2}^\text{ }^{2}p_{y2}^\text{ }^{2}p_{z2}^\text{ }^{3}s_{2}^\text{ }^{3}p_{x2}^\text{ }^{3}p_{y1}^\text{ }^{3}p_{z1}^\text{ }$

So,the number of orbitals involved is 9.

3 0

3 years ago

How much energy must be removed from a 94.4 g sample of benzene (molar mass= 78.11 g/mol) at 322.0 K to solidify the sample and

Kay [80]

Answer : The energy removed must be, 29.4 kJ

Explanation :

The process involved in this problem are :

$(1):C_6H_6(l)(322K)\rightarrow C_6H_6(l)(279K)\\\\(2):C_6H_6(l)(279K)\rightarrow C_6H_6(s)(279K)\\\\(3):C_6H_6(s)(279K)\rightarrow C_6H_6(s)(205K)$

The expression used will be:

$Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{fusion}]+[m\times c_{p,s}\times (T_{final}-T_{initial})]$

where,

$Q$ = heat released for the reaction = ?

m = mass of benzene = 94.4 g

$c_{p,s}$ = specific heat of solid benzene = $1.51J/g^oC=1.51J/g.K$

$c_{p,l}$ = specific heat of liquid benzene = $1.73J/g^oC=1.73J/g.K$

$\Delta H_{fusion}$ = enthalpy change for fusion = $-9.8kJ/mol=-\frac{9.8\times 1000J/mol}{78g/mol}=-125.6J/g$

Now put all the given values in the above expression, we get:

$Q=[94.4g\times 1.73J/g.K\times (279-322)K]+[94.4g\times -125.6J/g]+[94.4g\times 1.51J/g.K\times (205-279)K]$

$Q=-29427.312J=-29.4kJ$

Negative sign indicates that the heat is removed from the system.

Therefore, the energy removed must be, 29.4 kJ

3 0

3 years ago

In glycolysis, if glucose is labeled at the carbon 6 position (see page 1 for numbering of carbons in glucose) A) the carbon wit

Oliga [24]

Answer:

D) the carbon with the low-energy phosphate on it in 1,3 BPG is labeled.

Explanation:

Glycolysis has 2 phase (1) preparatory phase (2) pay-off phase.

<u>(1) Preparatory phase</u>

During preparatory phase glucose is converted into fructose-1,6-bisphosphate. Till this time the carbon numbering remains the same i.e. if we will label carbon at 6th position of glucose, its position will remian the same in fructose-1,6-bisphosphate that means the labeled carbon will still remain at 6th position.

When fructose-1,6-bisphosphate is further catalyzed with the help of enzyme aldolase it is cleaved into two 3 carbon intermediates which are glyceraldehyde 3-phosphate (GAP) and dihyroxyacetone phosphate (DHAP). In this conversion, the first three carbons of fructose-1,6-bisphosphate become carbons of DHAP while the last three carbons of fructose-1,6-bisphosphate will become carbons of GAP. It simply means that GAP will acquire the last carbon of fructose-1,6-bisphosphate which is labeled. Now the last carbon of GAP which has phosphate will be labeled.

<u>(2) Pay-off phase</u>

During this phase, GAP is dehydrogenated into 1,3-bisphosphoglycerate (BPG) with the help of enzyme glyceraldehyde 3-phosphate dehydrogenase. This oxidation is coupled to phosphorylation of C1 of GAP and this is the reason why 1,3-bisphosphoglycerate has phosphates at 2 positions i.e. at position 1 in which phosphate is newly added and position 3rd which already had labeled carbon.

It is pertinent to mention here that<u> BPG has a mixed anhydride and the bond at C1 is a very high energy bond.</u> In the next step, this high energy bond is hydrolyzed into a carboxylic acid with the help of enzyme phosphoglycerate kinase and the final product is 3-phosphoglycerate. Hence, the carbon with low energy phosphate i.e. the carbon at 3rd position remains labeled.

3 0

3 years ago

What equipment is needed to weigh 5.0 grams of zinc?

otez555 [7]

You would weigh the zinc with a weight, Zinc is a mineral so wether it's a solid or a liquid you would measure it with a zinc.

5 0

3 years ago

6. 5.50 x 10- molecules of carbon dioxide to moles.

sergeinik [125]

1.24973017189471 is probably the answer to your equation

4 0

3 years ago