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3 years ago

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if the density of a bar of gold is 19.3g/cm^3and you cut it into four pieces , what is its density of each piece of gold

1 answer:

AlladinOne [14]3 years ago

4 0

Density is an intrinsic property, so it is independent of the amount of substance present: one gold coin would have the same density as a solid gold boulder.

So if the density of gold is 19.3 g/cm³, the density of a bar of gold and the pieces into which the bar is cut would all be 19.3 g/cm³.

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The empirical formula of an organic compound is C2H4O. The molecular mass of the compound is 176g/mol.

Brrunno [24]

Answer:

The molecular formula of the compound is $C_{8}H_{16}O_{4}$ . The molecular formula is obtained by the following expression shown below

$\textrm{Molecular formula }= n\times \textrm{Empirical formula}$

Explanation:

Given molecular mass of the compound is 176 g/mol

Given empirical formula is $C_{2}H_{4}O$

Atomic mass of carbon, hydrogen and oxygen are 12 u , 1 u and 16 u respectively.

Empirical formula mass of the compound = $\left ( 2\times12+4+16 \right ) \textrm{ u} = 44 \textrm{ g/mol}$

$n = \displaystyle \frac{\textrm{Molecular formula mass}}{\textrm{Empirical formula mass}} \\n = \displaystyle \frac{176}{44} = 4$

$\textrm{Molecular formula }= n\times \textrm{Empirical formula}$

Molecular formula = 4 $\times$ $C_{2}H_{4}O$

Molecular formula is $C_{8}H_{16}O_{4}$

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a sealed container filled with argon gas at 35 c has a pressure of 832 torr. if the volume of the container is decreased by a fa

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Answer:

If the volume of the container is decreased by a factor of 2 the pressure is is increased by the same factor to 1664 torr.

Explanation:

Here we have Boyle's law which states that, at constant temperature, the volume of a given mass of gas is inversely proportional to its pressure

V ∝ 1/P or V₁·P₁ = V₂·P₂

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V₁ = Initial volume

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P₁ = Initial pressure = 832 torr

P₂ = Final pressure = Required

From V₁·P₁ = V₂·P₂ we have,

P₂ = V₁·P₁/V₂ = V₁·P₁/(V₁/2)

P₂ = 2·V₁·P₁/V₁ = 2·P₁ = 2× 832 torr = 1664 torr

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Explanation:

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Answer:

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The isomers of butane are

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The structures are shown in figure.

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