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Shkiper50 [21]
3 years ago
14

When a plant responds to changes in the length of day by flowering during certain seasons, it is called

Chemistry
2 answers:
lina2011 [118]3 years ago
8 0

Answer: photoperiodism

Zanzabum3 years ago
7 0

Answer:  photoperiodism

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The reaction in which hydrogen and oxygen are produced by running an electric current through water is an example of
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Decomposition is the  example  of a  reaction  in  which  hydrogen and oxygen  are  produced  by  running   an  electric current  through  water.

 Decomposition  of   water through  electrolysis  of  water  into   oxygen  and hydrogen   gas  due  an electric  current  passed  through   water.   when an  electric  current  is passed through  water, oxygen   gas is being  produced  at  the anode  and  hydrogen  gas  is produced  at the  cathode.
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3 years ago
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The environment can be changed in many ways, both by humans and nature. Some of these changes are good, while some are not. Sort
Ksju [112]

Answer:

Answers below (just some ethics)

Explanation:

Bad= Earthquake activity causing damage in a major city.

Bad= Producing and emitting smog from a smokestack.

Good= Planting native bushes to help control erosion.

Good= Volcanic eruption resulting in the formation of an island.

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What are miscible liquids?
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Explanation:

like water or ethanol

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Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang
Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

\Delta (H_{f})_{MnO_{2}} = -520.0 KJ/mole

\Delta (H_{f})_{Al_{2}O_{3}} = -1699.8 KJ/mole

The balanced chemical reaction is,

3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)

Formula used :

\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}

\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]

                        = (-3399.6) + (1560)

                        = -1839.6 KJ



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