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3 years ago

Light travels three-fourths as fast through water as it does through a vacuum.

Chemistry

2 answers:

Talja [164]3 years ago

4 0

1.33 is the answer thank me later

Marysya12 [62]3 years ago

4 0

The answer is 1.33hope it helps

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69. Sequencing Arrange these hydrates in order of

Kobotan [32]

In order of increasing percent water content:CoCl₂.6H₂O, Ba(OH)₂.8H₂O, MgSO₄.7H₂O

<h3>Further explanation</h3>

$\tt \%element=\dfrac{Ar~element}{MW~compound}\times 100\%$

CoCl₂.6H₂O.MW=237.90 g/mol

6H₂O MW = 6.18=108 g/mol

$\tt \%H_2O=\dfrac{108}{237.9}\times 100\%=45.4\%$

MgSO₄.7H₂O.MW=246.48 g/mol

MW 7H₂O = 7.18=126 g/mol

$\tt \%H_2O=\dfrac{126}{246.48}\times 100\%=51.1\%$

Ba(OH)₂.8H₂O MW=315.48 g/mol

MW 8H₂O = 8.18=144 g/mol

$\tt \%H_2O=\dfrac{144}{315.48}\times 100\%=45.6\%$

4 0

2 years ago

if the lightbulb receives 100 J of electrical energy, and gives off 75 energy, how much heat (thermal energy away from the light

oksano4ka [1.4K]

Answer:

Amount of heat energy released by light bulb = 25 joules

Explanation:

Given:

Energy receive by light bulb = 100 Joules

Energy released by light bulb as light energy = 75 Joules

Find:

Amount of heat energy released by light bulb

Computation:

We know that, energy is neither be created nor destroys

So,

Using Law of conservation of energy

Energy receive by light bulb = Energy released by light bulb as light energy + Amount of heat energy released by light bulb

100 = 75 + Amount of heat energy released by light bulb

Amount of heat energy released by light bulb = 100 - 75

Amount of heat energy released by light bulb = 25 joules

8 0

2 years ago

Help me with 23 fast

alina1380 [7]

Answer:

a salt

Explanation:

compounds are made of molecules but we need to mention the type of compound formed.

5 0

3 years ago

Read 2 more answers

: An unknown metal crystallizes in the cubic crystal structure. The metal has a radius 140pm, atomic mass of 135 g/mol, and dens

ki77a [65]

The cubic unit cell this metal crystallize as is BCC structure .

The structure of a crystalline solid, whether a metal or not, is best described by considering its simplest repeating unit, which is referred to as its unit cell.

The unit cell consists of lattice points that represent the locations of atoms or ions.

The entire structure then consists of this unit cell repeating in three dimensions

$\rm \rho =\dfrac{nM}{N_{0} a^{3}} \\\\\\\\\rm n =\dfrac{\rho N_{0} a^{3}}{M} \\\\\\\\Assuming\; it \; to \; be \; a\; BCC\; structure \\\\\\ r = \dfrac{\sqrt{3} \times a}{4} \\\\\\Therefore\; a = 3.23 \times 10^{-8}\\\\\\\\\rm n =\dfrac{13.3 \times 6.022 \times 10^{23}\times 3.23^{3}\times (10^{-8})^{3}}{135} \\\\$