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7nadin3 [17]
2 years ago
15

Please help with 4 and 5!

Chemistry
1 answer:
Rashid [163]2 years ago
5 0
B,c hop you git a good grad
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Please help me vote you brainiest
bekas [8.4K]

Explanation:

An object in motion <u>stays</u> <u>at</u> motion An object <u>at</u> <u>rest</u> stays at rest unless acted by an <u>external</u> force.

[ Newton's 1st law of motion ]

6 0
2 years ago
What are the stages of the life cycle of the sun
Jet001 [13]
The Sun is currently a main sequence star and will remain so for another 4-5 billion years. It will then expand and cool to become a red giant, after which it will shrink and heat up again to become a white dwarf. The white dwarf star will run out of nuclear fuel and slowly cool down over many billions of years.
5 0
2 years ago
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A penny has a mass of 2.50g and the Moon has a mass of ×7.351022kg, How many moles of pennies have a mass equal to the mass of t
baherus [9]
Answer: 122 moles

Procedure:

1) Convert all the units to the same unit

2) mass of a penny = 2.50 g

3) mass of the Moon = 7.35 * 10^22 kg (I had to arrage your numbers because it was wrong).

=> 7.35 * 10^22 kg * 1000 g / kg = 7.35 * 10^ 25 g.

4) find how many times the mass of a penny is contained in the mass of the Moon.

You have to divide the mass of the Moon by the mass of a penny

7.35 * 10^ 25 g / 2.50 g = 2.94 * 10^25 pennies

That means that 2.94 * 10^ 25 pennies have the mass of the Moon, which you can check by mulitiplying the mass of one penny times the number ob pennies: 2.50 g * 2.94 * 10^25 = 7.35 * 10^25.

5) Convert the number of pennies into mole unit. That is using Avogadros's number: 6.022 * 10^ 23

7.35 * 10^ 25 penny * 1 mol / (6.022 * 10^ 23 penny) = 1.22* 10^ 2 mole = 122 mol.

Answer: 122 mol
8 0
2 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

So  

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}

So  

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
How many sulfur atoms are present in 25.6 g of Al2(S2O3)3
IgorC [24]

Given the mass of Al_{2}(S_{2}O_{3})_{3}=25.6 g

The molar mass of Al_{2}(S_{2}O_{3})_{3}=390.35g/mol

Converting mass of Al_{2}(S_{2}O_{3})_{3}to moles:

25.6 g Al_{2}(S_{2}O_{3})_{3}*\frac{1molAl_{2}(S_{2}O_{3}}{390.35 gAl_{2}(S_{2}O_{3}} =0.0656molAl_{2}(S_{2}O_{3}

Converting mol Al_{2}(S_{2}O_{3})_{3}to mol S:

0.0656mol Al_{2}(S_{2}O_{3})_{3}*\frac{6molS}{1mol Al_{2}(S_{2}O_{3})_{3}}=0.3936 molS

Converting mol S to atoms of S using Avogadro's number:

1 mol = 6.022*10^{23}atoms

0.3936mol S *\frac{6.022*10^{23}atoms S}{1 mol S}=2.37*10^{23} S atoms

5 0
2 years ago
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