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2 years ago

How do you know when writing out equations the state of the element or compound

Chemistry

1 answer:

erica [24]2 years ago

7 0

Answer:

The state of matter is usually written in parentheses as a subscript directly after the element or compound.

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What is the percent composition by mass of chlorine in LiCI

Lelechka [254]

Answer: Lithium (Li) 16.373%

Chlorine (Cl) 83.627%

Explanation:

6 0

2 years ago

Need help on 17,19,20 please.

pentagon [3]

Answer:

c, maybe d, and I think b.

Explanation:

Im sorry if wrong

5 0

2 years ago

How do scientists know how atoms are structured?

astraxan [27]

Explanation:

i guess it's through thousands of research and experiments they conducted

5 0

2 years ago

Assume that the heat lost by the surface is gained by ethyl chloride. What enthalpies must you consider if you were to calculate

Ann [662]

Answer:

Specific heat of ethyl chloride in gas and liquid phases, enthalpy of vaporization and specific heat of solid surface.

Explanation:

In order to determine the final temperature, the heat lost by the chloride needs to be found. This would require the specific heat in both phases and the enthalpy of vaporization. (you will use q=mc(delta)T and q=m(delta)H)

Then the energy gained by the surface needs to be found. This will require the specific heat in order to use the q=mc(delta)T equation.

3 0

3 years ago

Fatty acids spread spontaneously on water to form a monomolecular film. A solution containing 0.10 mm^3 of a fatty acid is dropp

lara31 [8.8K]

Divide the volume by the area. Using scientific makes things a bit cleaner.

$0.10\,\mathrm{mm}^3 = 10^{-1}\,\mathrm{mm}^3$

$400.\,\mathrm{cm}^2 = 4\times10^2\,\mathrm{cm}^2$

Then

$\dfrac{10^{-1} \,\mathrm{mm}^3}{4\times10^2\,\mathrm{cm}^2} \cdot \dfrac{\left(\frac{1\,\rm m}{10^3\,\rm mm}\right)^3}{\left(\frac{1\,\rm m}{10^2\,\rm cm}\right)^2} = \dfrac{10^{-1}\times10^{-9} \,\mathrm m^3}{4\times10^2\times10^{-4}\,\mathrm m^2} = \dfrac{10^{-10}}{4\times10^{-2}}\,\mathrm m \\\\ ~~~~~~~~= 0.25\times10^{-8}\,\mathrm m$

Now, 1 m = 10⁹ nm, so

$0.25 \times10^{-8}\,\mathrm m \cdot \dfrac{10^9\,\rm nm}{1\,\rm m} = 0.25\times10^1\,\mathrm{nm} = \boxed{2.5\,\rm nm}$

8 0

1 year ago