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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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How many grams of magnesium oxide are needed to produce 264g of magnesium hydroxide
Archy [21]
If its one part magnesium and two parts hydroxide id say 88g of magnesium
3 0
3 years ago
1.Explain why the reactivity of group 7 decreases as you move down the group. Try to use the sentence starters here: When group
Kruka [31]

Answer: The reactivity of group 7 decreases as we move down the group because:

Explanation:

The elements of group 7 that is fluorine to iodine. The halogens are non metals and they react with metals to gain electrons. The metals loose electrons and the non metal gains it.

As we move down the group  the atomic radius gets bigger( more electron and more proton) and as a result the outer shells move further away from the nucleus.

There is more distance between the negatively charged electrons and positively charged nucleus.

Therefore the force of attraction between the shells and nucleus is lesser or weaker.

This makes attracting an extra electron from metals very difficult which results in weaker reaction.

Consequently, the reactivity decreases as we move down the group 7

5 0
3 years ago
One mechanism for the synthesis of ammonia proposes that N₂ and H₂ molecules catalytically dissociate into atoms:N₂ ⇆ 2N(g) log
Pavlova-9 [17]

The artificial fixation of nitrogen (N2) has enormous energy, environmental, and societal impact, the most important of which is the synthesis of ammonia (NH3) for fertilizers that help support nearly half of the world's population.

<h3>Artificial fixation of nitrogen</h3>

a) The equilibrium constant expression is Kp​=PCH4​​ PH2​ O​P CO​×PH 2​3​​.

(b) (i) As the pressure increases, the equilibrium will shift to the left so that less number of moles are produced.

(ii) For an exothermic reaction, with the increase in temperature, the equilibrium will shift in the backward direction.

(iii) When a catalyst is used, the equilibrium is not disturbed. The equilibrium is quickly attained

To learn more about equilibrium constant visit the link

brainly.com/question/10038290

#SPJ4

6 0
1 year ago
The standard enthalpy change for the following reaction is 873 kJ at 298 K.
Flura [38]

Answer:  - 436.5 kJ.

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation.

The given chemical reaction is,

2KCl(s)\rightarrow 2K(s)+Cl_2(g)  \Delta H_1=873kJ

Now we have to determine the value of \Delta H for the following reaction i.e,

K(s)+\frac{1}{2}Cl_2(g)\rightarrow KCl(s) \Delta H_2=?

According to the Hess’s law, if we divide the reaction by half then the \Delta H will also get halved and on reversing the reaction , the sign of enthlapy changes.

So, the value \Delta H_2 for the reaction will be:

\Delta H_2=\frac{1}{2}\times (-873kJ)

\Delta H_2=-436.5kJ

Hence, the value of \Delta H_2 for the reaction is -436.5 kJ.

8 0
3 years ago
Which of the following shows the conservation of mass during cellular respiration? 3 CO2 3 H2O energy → 3 C6H12O6 3 O2 6 CO2 6 H
Thepotemich [5.8K]

The reaction that has been, following the law of conservation of mass has been \rm C_6H_1_2O_6\;+\;6\;O_2\;\rightarrow\;6\;CO_2\;+\;6\;H_2O\;+\;Energy.

The law of conservation has been given in the chemical reaction that there has been no loss or gain of the mass and energy.

The law of conservation has been evident when there has been an equal number of atoms of each element on the product and the reactant side.

<h3 /><h3>Conservation of mass in Cellular respiration</h3>

The following reactions have been identified as:

  • \rm 3\;CO_2\;+\;3\;H_2O\;+\;Energy\;\rightarrow\;3\;C_6H_1_2O_6

Carbon atoms

Reactant = 3

Product = 18

Oxygen atoms

Reactant = 9

Product = 18

Hydrogen atoms

Reactant = 6

Product = 36

The number of atoms is not equal on the product and reactant side, thus not follows the law of conservation of mass.

  • \rm 3\;O_2\;+\;6\;CO_2\;+\;6\;H_2O\;+\;Energy\;\rightarrow\;C_6H_1_2O_6\;+\;6\;O_2

Carbon atoms

Reactant = 6

Product = 6

Oxygen atoms

Reactant = 24

Product = 18

Hydrogen atoms

Reactant = 12

Product = 12

The number of atoms is not equal on the product and reactant side, thus not follows the law of conservation of mass.

  • \rm C_6H_1_2O_6\;+\;6\;O_2\;\rightarrow\;6\;CO_2\;+\;6\;H_2O\;+\;Energy

Carbon atoms

Reactant = 6

Product = 6

Oxygen atoms

Reactant = 18

Product = 18

Hydrogen atoms

Reactant = 12

Product = 12

The number of atoms is equal on the product and reactant side, thus follows the law of conservation of mass.

  • \rm 6\;H_2O\;+\;C_6H_1_2O_6\;\rightarrow\;6\;O_2\;+\;Energy

Carbon atoms

Reactant = 6

Product = 0

Oxygen atoms

Reactant = 12

Product = 12

Hydrogen atoms

Reactant = 24

Product = 0

The number of atoms is not equal on the product and reactant side, thus not follows the law of conservation of mass.

Thus, the reaction that has been following the law of conservation of mass has been \rm C_6H_1_2O_6\;+\;6\;O_2\;\rightarrow\;6\;CO_2\;+\;6\;H_2O\;+\;Energy.

Learn more about law of conservation, here:

brainly.com/question/2175724

4 0
2 years ago
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