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damaskus [11]
1 year ago
13

Which are examples of biotic factors? Select three options.

Chemistry
1 answer:
alina1380 [7]1 year ago
6 0

The biotic factors are bacteria soil, dead leaves, and stream water.

<h3>What is an abiotic factor?</h3>

An abiotic factor is a non-living part of an ecosystem that shapes its environment.

Biotic and abiotic factors make up a community via interaction.

Biotic factors are considered living things (having "life") while abiotic factors are simply non-living things.

The dead leaves of plants are an abiotic factor, the bacteria in soil are living matter and stream flowing, etc.

Hence, the biotic factors are bacteria soil, dead leaves, and stream water.

Learn more about the abiotic factor here:

brainly.com/question/12689972

#SPJ1

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Breaking bonds requires what type of energy flow?
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Explanation:

the energy that that is needed to break a bond is called the bond energy or dissciation energy

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The weight of the body decrease inside water why?​
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What mass of oxygen gas is consumed in a reaction that produces 4.60 mol sulfur dioxide?
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7 0
2 years ago
The standard enthalpy of formation for glucose [c6h12o6(s)] is −1273.3 kj/mol. what is the correct formation equation correspond
balu736 [363]
The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is 
     C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is 
     6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)

Using the equation for the standard enthalpy change of formation 
     ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
     ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}

C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
     ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
                           = -1273.3 - (0 + 0 + 0)
                           = -1273.3
8 0
3 years ago
Some thermodynamic properties of ethanol are listed in the table. Thermodynamic Properties Property Value c (solid) 0.5 J/g °C c
alexira [117]
<h2>Answer </h2>

Option C - 320J

<u>Explanation </u>

Since ethanol solid at −120 °C and is only cooling down (it won’t change states) . The amount of Thermodynamic properties values c is given in form of solid, liquid and gas. Amount of energy released is calculated below.  

Formula,

= change in temperature  x specific heat capacity for solid ethanol x 40

=> 0.5 x 16x 40 = 320J

Therefore, the 320J of heat is released when 40.0g of ethanol cools.

4 0
3 years ago
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