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levacccp [35]
3 years ago
10

Suppose you conducted a similar experiment using helium gas. Given that helium is much lighter than the gases used in this lab,

what would you expect the relationship between the volume of helium and the pressure to be?
Chemistry
1 answer:
Anuta_ua [19.1K]3 years ago
8 0

This is an incomplete question, here is a complete question.

Suppose you conducted a similar experiment using helium gas. Given that helium is much lighter than the gases used in this lab, you would expect the relationship between the volume of helium and the pressure to be

A. periodic, because helium is lighter.

B. inversely proportional, just like propane and butane.

C. exponential, because helium is lighter.

D. directly proportional, just like propane and butane.

Answer : The correct option is, (B) inversely proportional, just like propane and butane.

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

That means, there is a inverse relation between volume of pressure.

From the given options we conclude that the correct statement is, inversely proportional, just like propane and butane.

Hence, the relationship between the volume of helium and the pressure to be inversely proportional, just like propane and butane.

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Answer: D) It has two more electrons than protons.

Explanation: Sulphur (S) is an element with atomic number 16 and thus contains 16 electrons. A neutral atom contains equal number of protons and electrons.

S:16:[Ne]3s^23p^4

An atom on losing electrons gains positive charge and on gaining electrons gain negative charge.

Sulphide ion (S^{2-}) is formed by gaining 2 electrons to attain stable configuration of argon and thus contains two more electrons than protons.

S^{2-}:18:[Ne]3s^23p^6:[Ar]

8 0
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A scientist wants to use a model to help present the results of his detailed scientific investigation
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because the model makes the concepts easier to understand

Explanation:

A model would be useful because they make concepts easier to understand.

Models are abstractions of the real world.

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6 0
3 years ago
A hydrated sample of cobalt II chloride has a molar mass of 309.99 g/mol. What is the mole ratio between cobalt II chloride and
VLD [36.1K]

Explanation:

Let there be 1 mole of CoCl2.

Mass of CoCl2 = 1mol * (129.839g/mol) = 129.839g

Mass of H2O = 309.99g - 129.839g = 180.151g

Moles of H2O = 180.151g / (18g/mol) = 10mol

Hence the mole ratio of CoCl2 to H2O is 1 : 10.

4 0
3 years ago
Read 2 more answers
Use Boyle's Law to explain what would happen to the volume of a helium- filled balloon if it was carried underwater by a diver t
oksano4ka [1.4K]

The deeper the diver takes the helium balloon, the more it reduces in size. This is due to the pressure of the water column above pressing on the balloon. According to Boyle’s law (P= k*1/V.), as the volume of the balloon decreases, the pressure of the helium inside increases.

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Q6. A student mixes 50.0 mL of 1.00 M Ba(OH)2 with 86.4 mL of 0.494 M H2SO4. Calculate the
skelet666 [1.2K]

Answer:

The correct answer should be 12.72 pH of the mixed solution and 9.94g mass of solid BASO4 formed.

Explanation:

The molecules of Ba(OH)2 are : 1.00M x 0.05L = 0.05 ( moles )

The molecules of H2SO4 are : 0.494M x 0.0864L = 0.0426816

Ba(OH)2 + H2SO4 ----> BaSO4 + 2 H2O

0.0426816<--- 0.0426816

The mixed solution is Ba(OH)2 with 0.05 - 0.0426816 = 0.0073184

The concentration of mixed solution is : 0.0073184 : ( 0.05 + 0.0864 ) = 0.054 M

The pH of mixed solution is 14 - -log[0.054] = 14 - 1.27 = 12.73 PH

And the mass of BaSO4 is 0.0426816 x ( 137 + 32 + 16 x 4 ) = 9.94 gams.

6 0
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