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1 year ago

7. A bottle found on a shelf contains either ethanol (density 0.789g/mL) or antifreeze (ethylene

Chemistry

1 answer:

loris [4]1 year ago

3 0

The substance is ethylene glycol.

<h3>What is density?</h3>

The term density is the ratio of mass to volume. We have two substances here which are ethanol and ethylene glycol. We have the mass of the solution as shown in the image in question 7 in the attachment to the question.

Mass of the solution = 381 - 250 = 131 g

Volume of the solution = 4 fl oz

1mL= 0.0338 fl

x = 4 fl oz

x =4 fl oz* 1mL/0.0338 fl oz

x = 118.34 mL

Density = mass/volume = 131 g/118.34 mL = 1.11 g/mL

The substance is ethylene glycol.

Learn more about density:brainly.com/question/15164682

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If a reaction occurs, what will be the products of the unbalanced reaction below?Zn(s) + HCl(aq) -->

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The products will be $ZnCl_2(s) + H_2(g)$

<h3>Chemical reactions</h3>

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2 years ago

Does every rock go through the rock cycle, from igneous to sedimentary to metamorphic each time around???

Alex17521 [72]

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3 years ago

A sample of gaseous neon atoms at atmospheric pressure and 0 °c contains 2.69 * 1022 atoms per liter. the atomic radius of neon

omeli [17]

Explanation

Radius of neon atom : 69 pm = $69\times 10^{-12} m$

Volume occupied by the one atom: $\frac{4}{3}\pi r^3$

$=\frac{4}{3}\times 3.14\times(69\times 10^{-12} m)^3=1.37\times 10^{-30} m^3$

given that $2.69\times 10^{22}$ atoms are present in 1L

1 L = 0.001 $m^3$

The volume occupied by the $2.69\times 10^{22}$ neon atoms

$2.69\times 10^{22}\times 1.37\times 10^{-30} m^3=3.68\times 10^{-8} m^3$

Fraction of volume occupied by the neon atom:

$=\frac{3.68\times 10^{-8} m^3}{0.001 m^3}=3.68\times 10^{-11} m^3$

$3.68\times 10^{-11} m^3$

The fraction of of volume occupied by the neon atom is very less than the 1 L which indicates the presence of large amount of empty space between the atoms of the gas.

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3 years ago

Element Y has two natural isotopes Y-63 (62.940 amu) and Y-65 (64.928 amu). Calculate the atomic mass of element Y, given the ab

djverab [1.8K]

Answer:

Average atomic mass = 63.553 amu.

Explanation:

Given data:

Abundance of Y-63 = 69.17%

Abundance of Y-65 = 100 - 69.17 = 30.83%

Atomic mass of Y-63 = 62.940 amu

Atomic mass of Y-65 = 64.928 amu

Atomic mass of Y = ?

Solution:

Average atomic mass= (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100

Average atomic mass= (62.940×69.17)+(64.928×30.83) /100

Average atomic mass = 4353.560 + 2001.730 / 100

Average atomic mass = 6355.29 / 100

Average atomic mass = 63.553 amu.

5 0

3 years ago