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alexgriva [62]
3 years ago
9

Describe the energy inputs and outputs for the campfire. Use the law of conservation of energy to construct a valid qualitative

equation that includes all the input and output energies involved.
Chemistry
1 answer:
solniwko [45]3 years ago
5 0

I did some fire simulation work recently where a small paper fire (think like half a ream of printer paper) was putting out 4000 watts, or about 5 horsepower. A large roaring campfire is going to be a few multiples of that depending on size, of course.

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The Lanthanides are often be found below the Periodic Table so it will fit nicely on a textbook page. If they were inserted into
Annette [7]

Answer:

period 6

Explanation:

If the lanthanides were inserted into the periodic table, they would go into periodic 6.

Their atomic number is between 57 - 71 from element lanthanum to lutetium.

  • The elements in this period are 15 in number.
  • They are also know as elements in the f-block.

The elements that makes up the series are:

Lanthanum

Cerium

Praseodymium

Neodymium

Promethium

Samarium

Europium

Gadolinium

Terbium

Dysprosium

Holmium

Erbium

Thulium

Ytterbium

Lutetium

6 0
3 years ago
Which of the following is the balanced equation for the
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A.) is the answer =)
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3 years ago
A fusion reaction takes place between carbon and another element neutrons are released and a different element is formed the dif
Andreas93 [3]

A fusion reaction takes place between carbon and another element. Neutrons are released, and a different element is formed. The different element is Lighter than helium.

7 0
3 years ago
Do you think a single gene controls height in humans, as it does in peas? Explain your answer.
lakkis [162]
No, because humans are much more complex than peas.
6 0
3 years ago
Read 2 more answers
Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,
WITCHER [35]

Answer: Enthalpy of combustion (per mole) of C_4H_{10} (g) is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]

We are given:

\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ

Enthalpy of combustion (per mole) of C_4H_{10} (g) is -2657.5 kJ

6 0
2 years ago
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