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3 years ago

11

Which objects in space formed from the huge disk of ice and debris beyond the outer planets?

2 answers:

Oliga [24]3 years ago

5 0

Answer: comet

Explanation:

A comet is a body found in the outskirts of the solar system <u>in Kuiper belt or Oort clouds beyond the orbit of Neptune</u><u>. </u>A comet has <u>core </u>which rocky formed<u> from the debris of the solar system</u>. The core is known as nucleus which is surrounded by atmosphere of frozen gases. Comets orbit the Sun in elliptical path and enter the system when their path is perturbed by the disturbance in nearby star.

olga_2 [115]3 years ago

3 0

Comets are usually formed of ice and other suave debris, while asteroids typically contain metals.

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A single atom of platinum has a mass of 3.25 x 10-22g. What is the mass of 6.0 x 1023 platinum atoms?

boyakko [2]

Even though the numbers are in scientific notation, they are still just numbers. Essentially the question is as simple as "An apple weighs an ounce, how much do 35 apples weigh?". So all you have to do is multiply.

3.25 x $10^{-22}$ x 6.0 x $10^{23}$ = 195

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2 years ago

We have air (21% O2 and 79% N2) at 23 bar and 30 C. 4. What is the ideal molar volume (m^3/kmol)? a. b. What is the Z factor? Wh

k0ka [10]

Answer:

The ideal molar volume is $\frac{V}{n} =V_z= 0.001095 \ m^3/mol$

The Z factor is $Z = 0.09997$

The real molar volume is $\frac{V_r}{n} = V_k= 0.0001095\ \frac{m^3}{mol}$

Explanation:

From the question we are told that

The pressure is $P = 23 \ bar = 23 *10^5 Pa$

The temperature is $T = 30 ^ oC = 303 \ K$

According to the ideal gas equation we have that

$PV = nRT$

=> $\frac{V}{n}=V_z= \frac{RT}{P}$

Where $\frac{V}{n }$ is the molar volume and R is the gas constant with value

$R = 8.314 \ m^3 \cdot Pa \cdot K^{-1}\cdot mol^{-1}$

substituting values

$\frac{V}{n} =V_z= \frac{ 8.314 * 303}{23 *10^{5}}$

$\frac{V}{n} =V_z= 0.001095 \ m^3/mol$

The compressibility factor of the gas is mathematically represented as

$Z = \frac{P * V_z}{RT}$

substituting values

$Z = \frac{23 *10^{5} * 0.001095}{8.314 * 303}$

$Z = 0.09997$

Now the real molar volume is evaluated as

$\frac{V_r}{n} = V_k= \frac{Z * RT }{P}$

substituting values

$\frac{V_r}{n} = V_k= \frac{0.09997 * 8.314 * 303}{23 *10^{5}}$

$\frac{V_r}{n} = V_k= 0.0001095\ \frac{m^3}{mol}$

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