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Feliz [49]
3 years ago
7

Is it possible for a balloon with an initial pressure of 200.0 kPa to naturally expand to four times its initial volume when the

temperature remains constant and atmospheric pressure is 101.3kPa?
Chemistry
1 answer:
seropon [69]3 years ago
3 0
Yes it is possible for a balloon to expand under these conditions. When objects are exposed to an atmospheric pressure beyond their capabilities, their physical characteristics change. This is because, molecules within the balloon are constantly moving and reacting to its external environment.
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Pls HELP!!!! Why does copper oxide appear on a penny faster when a penny is heated in a flame than when the penny is at room tem
balu736 [363]

The rate of reaction between oxygen and copper is higher at high temperature. The rate of reaction between oxygen and copper is lower at room temperature.

at high temperature copper penny will react with oxygen as shown below

2Cu(s)+O_{2}-->2CuO

The reaction does not occur easily at room temperature

We can say that the reaction between copper and oxygen is an endothermic reaction so the rate of reaction is high at high temperature.

The bond energy of oxygen molecules need energy to react with copper and form copper oxide.

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The answer to question is C
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The cl-cl bond will react more readily, because it has a
bekas [8.4K]

Answer: bond energy

Explanation:

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Calcium has a cubic closest packed structure as a solid. Assuming that calcium has an atomic radius of 197 pm, calculate the den
NISA [10]

Answer:

\rho=1.54\ g/cm^3

Explanation:

The expression for density is:

\rho=\frac {Z\times M}{N_a\times {{(Edge\ length)}^3}}

N_a=6.023\times 10^{23}\ {mol}^{-1}

M is molar mass of Calcium = 40.078 g/mol

For cubic closest packed structure , Z= 4

\rho is the density

Radius = 197 pm = 1.97\times 10^{-8}\ cm

Also, for fcc, Edge\ length=2\sqrt{2}\times radius=2\sqrt{2}\times 1.97\times 10^{-8}\ cm=5.572\times 10^{-8}\ cm

Thus,  

\rho=\frac{4\times \:40.078}{6.023\times \:10^{23}\times \left(5.572\times 10^{-8}\right)^3}\ g/cm^3

\rho=\frac{160.312}{10^{23}\times \:6.023\left(10^{-8}\times \:5.572\right)^3}\ g/cm^3

\rho=\frac{160.312}{10^{23}\times \:1.04195E-21}\ g/cm^3

\rho=\frac{160.312}{104.19483}\ g/cm^3

\rho=1.54\ g/cm^3

7 0
3 years ago
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