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3 years ago

Standard units of measure for density

Chemistry

1 answer:

Mrrafil [7]3 years ago

8 0

Answer:

g/cm³ for solids,

g/ml for liquids

g/L for gases.

Explanation:

Though SI unit of density is kg/m³, for convenience we use g/cm³ for solids, g/ml for liquids and g/L for gases. Mathematically, density is defined as mass divided by volume: ρ=m/V

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A Lewis acid is a substance that can

katrin2010 [14]

Accept a pair of nonbonding electrons,a Lewis acid is an electron-pair acceptor. A Lewis base is any substance, such as the OH- ion, that can donate a pair of nonbonding electrons. A Lewis base is therefore an electron-pair donor.

6 0

3 years ago

Complete combustion of 2.60 g of a hydrocarbon produced 8.46 g of co2 and 2.60 g of h2o. what is the empirical formula for the h

zubka84 [21]

1 mole of carbon dioxide contains a mass of 44 g, out of which 12 g are carbon.
Hence, in this case the mass of carbon in 8.46 g of CO2:
(12/44) × 8.46 = 2.3073 g
1 mole of water contains 18 g, out of which 2 g is hydrogen;
Therefore, 2.6 g of water contains;
(2/18) × 2.6 = 0.2889 g of hydrogen.
Therefore, with the amount of carbon and hydrogen from the hydrocarbon we can calculate the empirical formula.
We first calculate the number of moles of each,
Carbon = 2.3073/12 = 0.1923 moles
Hydrogen = 0.2889/1 = 0.2889 moles
Then, we calculate the ratio of Carbon to hydrogen by dividing with the smallest number value;
Carbon : Hydrogen
0.1923/0.1923 : 0.2889/0.1923
1 : 1.5
(1 : 1.5) 2
= 2 : 3
Hence, the empirical formula of the hydrocarbon is C2H3

5 0

3 years ago

Convert 1.5 km to millimeters, and express the result in scientific notation.

Grace [21]

The answer is 1.5e+6. Hope this helped!

7 0

3 years ago

I NEED HELP PLEASEEEEEE I WILL RATE BRAINLIEST!!!!

MrRa [10]

0.002 moles is the answer

8 0

3 years ago

a) Whatis the composition in mole fractions of a solution of benzene and toluene that has a vapor pressure of 35 torr at 20 °C?

iogann1982 [59]

Answer:

molar composition for liquid

xb= 0.24

xt=0.76

molar composition for vapor

yb=0.51

yt=0.49

Explanation:

For an ideal solution we can use the Raoult law.

Raoult law: in an ideal liquid solution, the vapor pressure for every component in the solution (partial pressure) is equal to the vapor pressure of every pure component multiple by its molar fraction.

For toluene and benzene would be:

$P_{B}=x_{B}*P_{B}^{o}$

$P_{T}=x_{T}*P_{T}^{o}$

Where:

$P_{B}$ is partial pressure for benzene in the liquid

$x_{B}$ is benzene molar fraction in the liquid

$P_{B}^{o}$ vapor pressure for pure benzene.

The total pressure in the solution is:

$P= P_{T}+ P_{B}$

And

$1=x_{B}+x_{T}$

Working on the equation for total pressure we have:

$P=x_{B}*P_{B}^{o} + x_{T}*P_{T}^{o}$

Since $x_{T}=1-x_{B}$

$P=x_{B}*P_{B}^{o} + (1-x_{B})*P_{T}^{o}$

We know P and both vapor pressures so we can clear $x_{B}$ from the equation.

$x_{B}=\frac{P- P_{T}^{o}}{ P_{B}^{o} - P_{T}^{o}}$

$x_{B}=\frac{35- 22}{75-22} = 0.24$

$x_{T}=1-0.24 = 0.76$

To get the mole fraction for the vapor we know that in the equilibrium:

$P_{B}=y_{B}*P$

$y_{T}=1-y_{B}$

$y_{B} =\frac{P_{B}}{P}=\frac{ x_{B}*P_{B}^{o}}{P}$

$y_{B}=\frac{0.24*75}{35}=0.51$

$y_{T}=1-0.51=0.49$

Something that we can see in these compositions is that the liquid is richer in the less volatile compound (toluene) and the vapor in the more volatile compound (benzene). If we take away this vapor from the solution, the solution is going to reach a new state of equilibrium, where more vapor will be produced. This vapor will have a higher molar fraction of the more volatile compound. If we do this a lot of times, we can get a vapor that is almost pure in the more volatile compound. This is principle used in the fractional distillation.

7 0

3 years ago