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3 years ago

If 468mL of gas is measured at 0.83atm, then what volume would the gas occupy at 1.05atm?

Chemistry

1 answer:

Serggg [28]3 years ago

7 0

Answer: V= 369.94mL

Explanation:

Applying Boyle's law

P1V1=P2V2

P1= 0.83atm, V1= 468mL , P2 = 1.08atm V2= ?

Substitute into above

0.83× 468= 1.08× V2

V2 = 369.9mL

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bixtya [17]

Answer:

Factor 1 is a human induced factor and Factor 2 may be influenced by both nature and human factors.

Explanation:

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3 years ago

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Write the balanced chemical equations for the decomposition of solid calcium hydroxide into solid calcium (ii) oxide (lime) and

Lesechka [4]

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3 years ago

Mass = 200.0 g volume = 100.0cm3 What is the density?

Marianna [84]

The density of an object is defined as its mass divided by its volume. Mathematically, density = Mass / Volume. The unit of density is kilogram per cubic meter, kg / m^3 or g /cm^3.
For the question given above: the
Mass = 200.0 g
Volume = 100.0 cm^3
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3 years ago

Using the systematic approach for equilibrium problems, calculate the pH of 0.05 M HOCl. Ka= 3.0*10-8 Group of answer choices 3.

SVEN [57.7K]

Answer:

The pH is equal to 4.41

Explanation:

Since HClO is a weak acid, its dissociation in aqueous medium is:

HClO ⇄ ClO- + H+

start: 0.05 0 0

change -x +x +x

balance 0.05-x x x

As it is a weak acid it dissociates very little, in its ClO- and H + ions, so the change is negative, where x is a degree of dissociation.

the acidity constant when equilibrium is reached is equal to:

$Ka=\frac{[ClO-]*[H+]}{[HClO]}=\frac{x*x}{0.05-x}=3x10^{-8}$

The 0.05-x fraction can be approximated to 0.05, because the ionized fraction (x) is very small, therefore we have:

$3x10^{-8}=\frac{x^{2} }{0.05}$

clearing the x and calculating its value we have:

$x=3.87x10^{-5}=[H+]=[ClO-]$

the pH can be calculated by:

$pH=-log[H+]=-log[3.87x10^{-5}]=4.41$

7 0

3 years ago

Consider the following four structures: What is the relationship of I and II? What is the relationship of I and III? What is the

daser333 [38]

Answer:

The different structures are shown in the attachment.

I and II - structural isomers

I and III - Structural isomers

I and IV - structural isomers

II and III - structural isomers

II and IV - structural isomers

III and IV - stereoisomers

Explanation:

The knowledge of Isomerism is tested here; there are two types of isomerism ; structural and stereoisomerism.

Structural Isomers have similar molecular and different double bond positioning, these occurs mostly in ALKENE FAMILY.

Stereo-isomers have the same molecular formular and similar patterns but differ in their spatial arrangement. trans and cis are typical examples of stereo-isomers.

From the question; Relationship between I and II is that they are structural isomers since they have the same molecular formula, but different bond atom arrangement and infact they are the same compound.

Relationship between I and III is that they are structural isomers with similar molecular formular but differ in the double bond position.

Relationship between I and IV is that they are structural isomers with similar molecular formula but different double bond arrangement.

Relationship between II and III is that they are structural isomers with similar molecular formular but different double bond position

Relationship between II and IV is that they are also structural isomers with the same molecular formular but different double bond position.

Relationship between III and IV is that they are stereo-isomers with same molecular formula but different spatial arrangement, hence cis and trans.

4 0

3 years ago