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3 years ago

3. What form is energy in when it enters a food chain and when it leaves a food chain?

1 answer:

3 0

Hello Shobaravi6542, Im sorry that i am answering late.

When it enters the food chain, it is in the form of light from the sun! (From the sun) hen it leaves it is in the form of heat.

Glad i could help!

You might be interested in

Why do you think that tsunamis occur most commonly in the Pacific Ocean?

zhenek [66]

Tsunamis are most common in the Pacific Ocean Basin. The entire Pacific Ocean is ringed by areas known as subduction zones, where the tectonic plates of the earth are moving relative to one another. Subduction zones are where two or more plates are colliding and one is going underneath another.

5 0

3 years ago

Given the following data: Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = −23 kJ 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ΔH = −39

nalin [4]

Answer:

ΔH° = -11 kj

Explanation:

Step 1: Data given

1) Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = −23 kJ

2) 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ΔH = −39 kJ

3) Fe3O4(s) + CO(g) → 3 FeO(s) + CO2(g) ΔH = +18 kJ

Step 2: The balanced equation

FeO + CO → Fe + CO2

Step 3: Calculate ΔH for the reaction FeO(s) + CO(g) → Fe(s) + CO2(g).

To get this equation, we need to combine the 3 equations

We have to multiply the third equation by 2.

2Fe3O4(s) + 2CO(g) → 6 FeO(s) + 2CO2(g) ΔH = +54 kJ

This equation we add to the second equation

3Fe2O3(s) + CO(g) + 2Fe3O4(s) + 2CO(g) → 2Fe3O4(s) + CO2(g) + 6FeO(s) + 2CO2 (g)

3Fe2O3(s) + 3CO(g) → 3CO2(g) + 6FeO(s)

ΔH° = 2*18 + (-39) = 36 - 39 = -3 kJ

This new equation we will divide by 3

3Fe2O3(s) + 3CO(g) → 3CO2(g) + 6FeO(s)

Fe2O3(s) + CO(g) → CO2(g) + 2FeO(s)

ΔH° =-3/3 = -1 kJ

Now we will substract this new equation from the first equation

Fe2O3(s) + 3CO(g) - Fe2O3(s) - CO(g) → 2Fe(s) + 3CO2(g) -2FeO(s) - CO2(g)

2CO(g) + 2FeO(s) → 2Fe(s) + 2CO2(g)

ΔH° = -23kJ +1kJ

ΔH° = -22 kj

The next equation we will divide by 2

2CO(g) + 2FeO(s) → 2Fe(s) + 2CO2(g)

CO(g) + FeO(s) → Fe(s) + CO2(g)

ΔH° = -22kJ /2

ΔH° = -11 kj

7 0

3 years ago

Concentrated nitric acid is 15.6M and has a density of 1.41g/mL. What is the weight percent of concentrated HNO3?

tankabanditka [31]

Molarity= mol/ liters

since the molarity is given, we can assume that we have 1.0 Liters of solution

15.6 M= mol/ 1 liters---> this means that we have 15.6 moles of HNO3

we need to convert these moles to grams using the molar mass of HNO3

molar mass HNO3= 1.01 + 14.0 + (3 X 16.0)= 63.01 g/mol

15.6 mol HNO3 (63.01 g/ mol)= 983 grams HNO3

now we have to determine the grams of solution using the assumption of 1 liters of solution and the density

1 liters= 1000 mL

1000 mL (1.41 g/ ml)= 1410 grams solution

mass percent= mass of solute/ mass of solution x 100

mass percent= 63.01/ 1410 x 100= 4.47 %

6 0

4 years ago

Consider an insulated container in which water (2.58 g) solidifies at 0 oC and 1 atm on the surface of a 55 g aluminum block. As

yarga [219]

Answer:

17.41°C is the change in temperature of the aluminum block.

Explanation:

Latent heat of fusion water = 334 J/g

Heat required to freeze 1 gram of water = -334 J

Heat required to freeze 2.58 grams of water = Q

$Q=2.58 g\times (-334 J/g)=-861.72 J$