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4vir4ik [10]
3 years ago
10

What is the powerhouse of the cell?​

Chemistry
2 answers:
Rasek [7]3 years ago
7 0

Answer:

Mitochondria

Explanation:

Mitochondria is the powerhouse of the cell, convert sustenance into energy, fueling the cell's activities. In addition to power, mitochondria also produce reactive oxygen

blsea [12.9K]3 years ago
4 0
Mitochondria, the powerhouse of the cell, convert sustenance into energy, fueling the cell's activities. In addition to power, mitochondria also produce reactive oxygen species, byproduct molecules primed to help facilitate communication among the other units in the cells.
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In a college of exactly 2940 students, exactly 60 % are male. What is the number of female students?
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Answer:1176

Explanation:

Number of male students: 2940*(60/100)=1764

So the number of female students would be:

2940-1764=1176

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What is true of a compound?
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Answer:

It does not always retain the properties of the substances that make it up

Explanation:

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Explain the term 'end point'?<br> What do you mean by a 1.0 M solution?
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Answer:

According to me end point is the end of a reaction

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2 years ago
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How much energy must be removed from a 125 g sample of benzene (molar mass= 78.11 g/mol) at 425.0 K to liquify the sample and lo
riadik2000 [5.3K]

Answer : The energy removed must be, -67.7 kJ

Solution :

The process involved in this problem are :

(1):C_6H_6(g)(425.0K)\rightarrow C_6H_6(g)(353.0K)\\\\(2):C_6H_6(g)(353.0K)\rightarrow C_6H_6(l)(353.0K)\\\\(3):C_6H_6(l)(353.0K)\rightarrow C_6H_6(l)(335.0K)

The expression used will be:

\Delta H=[m\times c_{p,g}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,l}\times (T_{final}-T_{initial})]

where,

\Delta H = heat released by the reaction = ?

m = mass of benzene = 125 g

c_{p,g} = specific heat of gaseous benzene = 1.06J/g^oC

c_{p,l} = specific heat of liquid benzene = 1.73J/g^oC

\Delta H_{vap} = enthalpy change for vaporization = 33.9kJ/mole=33900J/mole=\frac{33900J/mole}{78.11g/mole}J/g=434.0J/g

Molar mass of benzene = 78.11 g/mole

Now put all the given values in the above expression, we get:

\Delta H=[125g\times 1.06J/g.K\times (353.0-(425.0))K]+125g\times -434.0J/g+[125g\times 1.73J/g.K\times (335.0-353.0)K]

\Delta H=-67682.5J=-67.7kJ

Therefore, the energy removed must be, -67.7 kJ

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2 years ago
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