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Lubov Fominskaja [6]

3 years ago

12

In science what can only be separated by Chemical methods

1 answer:

PilotLPTM [1.2K]3 years ago

5 0

Answer:

A mixture is a physical combination of substances thus it only requires physical processes to separate.

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Li Ming thinks that chlorophyll is a reactant of photosynthesis, but her classmate Julian disagrees. Who is correct, and why?

Eddi Din [679]

Julian is correct because chlorophyll is neither used up nor formed in the chemical reaction of photosynthesis.

Explanation:

Julian is correct because chlorophyll is neither used up nor formed in the chemical reaction of photosynthesis. The reactants in photosynthesis are carbon dioxide and water.

Photosynthesis is the process by which green plants manufacture their food in the presence of sunlight.
In photosynthesis carbon dioxide combines with water to produce glucose and oxygen.
Chlorophyll in plant is a green pigment that is used to trap sunlight during photosynthesis.

Learn more:

Photosynthesis brainly.com/question/2761166

#learnwithBrainly

7 0

3 years ago

Read 2 more answers

Write the half-reactions as they occur at each electrode and the net cell reaction for this electrochemical cell containing indi

AlexFokin [52]

Explanation:

The given cell reaction is as follows.

$In(s)| In^{3+}(aq) || Cd^{2+}(aq) | Cd(s)$

Hence, reactions taking place at the cathode and anode are as follows.

At anode ; Oxidation-half reaction : $In(s) \rightarrow In^{3+}(aq) + 3e^{-}$ ...... (1)

At cathode; Reduction-half reaction : $Cd^{2+}(aq) + 2e^{-} \rightarrow Cd(s)$ ....... (2)

Hence, balance the half reactions by multiplying equation (1) by 2 and equation (2) by 3.

Therefore, net cell reaction is as follows.

$2In(s) \rightarrow 2In^{3+}(aq) + 6e^{-}$

$3Cd^{2+}(aq) + 6e^{-} \rightarrow 3Cd(s)$

Net reaction: $2In(s) + 3Cd^{2+}(aq) \rightarrow 2In^{3+}(aq) + 3Cd(s)$

Thus, we can conclude that the overall cell reaction is as follows.

$2In(s) + 3Cd^{2+}(aq) \rightarrow 2In^{3+}(aq) + 3Cd(s)$

4 0

3 years ago

Name all nonmetals that are noble has.<br> PLEASE HELP ASAP ILL GIVE YOU BRAINLIEST

miss Akunina [59]

Hellium argon neon xenon krypton randon oxygen fluorine chlorine bromine

4 0

2 years ago

Base changes phenolphthalein to pink it is true or false

galben [10]

Answer:

It’s false

Explanation:

it could be true if the question mentioned alkaline solution

4 0

3 years ago

Read 2 more answers

Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h

bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

$\Delta S=-n\times C_v\ln \frac{T_2}{T_1}$

We know that,

The relation between the $C_p\text{ and }C_v$ for an ideal gas are :

$C_p-C_v=R$

As we are given :

$C_p=28.253J/K.mole$

$28.253J/K.mole-C_v=8.314J/K.mole$

$C_v=19.939J/K.mole$

Now we have to calculate the entropy change of the gas.

$\Delta S=-n\times C_v\ln \frac{T_2}{T_1}$

$\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J$

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. $(w=-pdV)$

(C) Heat during the process will be,

$q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J$

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

7 0

3 years ago