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3 years ago

5

Erosion and weathering can both cause changes to the surface of Earth. Which of the following happens only because of erosion an

d NOT because of weathering?

1 answer:

dangina [55]3 years ago

7 0

Answer:

caves

Explanation:

BRAINLIEST

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Calculate the density of air at 100 Deg C and 1 bar abs. Use the Ideal Gas Law for your calculation and give answer in kg/m3. Us

madam [21]

Answer:

$d=0.92\frac{kg}{m^{3}}$

Explanation:

Using the Ideal Gas Law we have $PV=nRT$ and the number of moles n could be expressed as $n=\frac{m}{M}$ , where m is the mass and M is the molar mass.

Now, replacing the number of moles in the equation for the ideal gass law:

$PV=\frac{m}{M}RT$

If we pass the V to divide:

$P=\frac{m}{V}\frac{RT}{M}$

As the density is expressed as $d=\frac{m}{V}$ , we have:

$P=d\frac{RT}{M}$

Solving for the density:

$d=\frac{PM}{RT}$

Then we need to convert the units to the S.I.:

$T=100^{o}C+273.15$

$T=373.15K$

$P=1bar*\frac{0.98atm}{1bar}$

$P=0.98atm$

$M=28.9\frac{kg}{kmol}*\frac{1kmol}{1000mol}$

$M=0.0289\frac{kg}{mol}$

Finally we replace the values:

$d=\frac{(0.98atm)(0.0289\frac{kg}{mol})}{(0.082\frac{atm.L}{mol.K})(373.15K)}$

$d=9.2*10^{-4}\frac{kg}{L}$

$d=9.2*10^{-4}\frac{kg}{L}*\frac{1L}{0.001m^{3}}$

$d=0.92\frac{kg}{m^{3}}$

5 0

3 years ago

How does kinetic energy affect the stopping distance of a vehicle traveling at 30 mph compared to the same vehicle traveling at

Anastasy [175]

The larger the kinetic energy of the vehicle, the larger the amount of energy will be needed to stop the vehicle, meaning that faster vehicles have a larger stopping distance

5 0

2 years ago

Determine the number of atoms that are in 1.15 mol of Cl 2.

Brums [2.3K]

Answer:

Explanation: I think its 4.91 x 10^25. Im not very sure, i just multipled 1.15 mol by the molar mass of Cl 2, which was 70.9 g. Then I multiplied that by avogadro's number. sorry if im wrong

4 0

2 years ago

The vapor pressure of water at 65oC is 187.54 mmHg. What is the vapor pressure of a ethylene glycol (CH2(OH)CH2(OH)) solution ma

Pavlova-9 [17]

Answer:

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

Explanation:

Vapor pressure of water at 65 °C= $p_o= 187.54 mmHg$

Vapor pressure of the solution at 65 °C= $p_s$

The relative lowering of vapor pressure of solution in which non volatile solute is dissolved is equal to mole fraction of solute in the solution.

Mass of ethylene glycol = 22.37 g

Mass of water in a solution = 82.21 g

Moles of water= $n_1=\frac{82.21 g}{18 g/mol}=4.5672 mol$

Moles of ethylene glycol= $n_2=\frac{22.37 g}{62.07 g/mol}=0.3603 mol$

$\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}$

$\frac{187.54 mmHg-p_s}{187.54 mmHg}=\frac{0.3603 mol}{0.3603 mol+4.5672 mol}$

$p_s=173.83 mmHg$

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

6 0

2 years ago

Give the number of significant figures: 369,132,000 sec

Yuliya22 [10]

The number of significant figures in 369,132,000 is 6

3 0

2 years ago

Read 2 more answers