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3 years ago

Estimate the volume of a piece of molecular cloud that has the same amount of water as your body.

Physics

1 answer:

noname [10]3 years ago

6 0

Question:

The water molecules now in your body were once part of a molecular cloud. Only about onemillionth of the mass of a molecular cloud is in the form of water molecules, and the mass density of such a cloud is roughly 2.0×10−21 g/cm^3.

Estimate the volume of a piece of molecular cloud that has the same amount of water as your body.

Answer:

The volume of cloud that has the same density as the amount of water in our body is 1.4×10²⁵ cm³

Explanation:

Here, we have mass density of cloud = 2.0×10⁻²¹ g/cm^3

Density = Mass/Volume

Volume = Mass/Density = If the mass is 40 kg and the body is made up of 70% by mass of water, we have

28 kg water = 28000 g

Therefore the Volume = 28 kg/ 2.0×10⁻²¹ g/cm^3 = 1.4×10¹⁹ m³ = 1.4×10²⁵ cm³.

Therefore, the volume of cloud that has the same density as the amount of water in our body = 1.4×10²⁵ cm³.

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Musya8 [376]

Answer:

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2 years ago

Read 2 more answers

If the concentration of Ag+ is 0.0115 M, the concentration of H+ is 0.355 M, and the pressure of H2 is 1.00 atm, calculate the c

scoundrel [369]

Answer: The cell potential is 0.712 V

Explanation:

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, silver will undergo reduction reaction will get reduced. Hydrogen will undergo oxidation reaction and will get oxidized

Half reactions for the given cell follows:

Oxidation half reaction: $H_2(1.00atm)\rightarrow 2H^{+}(0.355M)+2e^-;E^o_{H^{+}/H_2}=0.00V$

Reduction half reaction: $Ag^{+}(0.0115M)+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V$ ( × 2 )

Net reaction: $H_2(1.00atm)+2Ag^{+}(0.0115M)\rightarrow 2H^{+}(0.355M)+2Ag(s)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.80-(0.00)=0.80V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^2}{[Ag^{+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = ?V

$E^o_{cell}$ = standard electrode potential of the cell = +0.80 V

n = number of electrons exchanged = 2

$[Ag^{+}]=0.0115M$

$[H^{+}]=0.355M$

Putting values in above equation, we get:

$E_{cell}=0.80-\frac{0.059}{2}\times \log(\frac{(0.355)^2}{(0.0115)^2})\\\\E_{cell}=0.712V$

Hence, the cell potential is 0.712 V

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3 years ago

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marta [7]

Answer:

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Explanation:

When an atom is electrically polarized, it means that one side has more electrons than the other. This makes one side more negative as it has more electrons than the other. I hope this helps!

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