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3 years ago

How do the orbits of a comet and earth compare?

1 answer:

8 0

Comets are balls of ice and dust in orbit around the Sun. The orbits of comets are different from those of planets - they are elliptical. A comet's orbit takes it very close to the Sun and then far away again.

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A student weighs an empty flask and stopper and finds the mass to be 55.844 g. She then adds about 5 mL of an unknown liquid and

Oduvanchick [21]

Answer :

(a) The pressure of the vapor in the flask in atm is, 0.989 atm

(b) The temperature of the vapor in the flask in Kelvin is, 372.7 K

The volume of the flask in liters is, 0.2481 L

(c) The mass of vapor present in the flask was, 0.257 g

(d) The number of moles of vapor present are 0.00802 mole.

(e) The mass of one mole of vapor is 32.0 g/mole

Explanation : Given,

Mass of empty flask and stopper = 55.844 g

Volume of liquid = 5 mL

Temperature = $99.7^oC$

Mass of flask and condensed vapor = 56.101 g

Volume of flask = 248.1 mL

Barometric pressure in the laboratory = 752 mmHg

(a) First we have to determine the pressure of the vapor in the flask in atm.

Pressure of the vapor in the flask = Barometric pressure in the laboratory = 752 mmHg

Conversion used :

$1atm=760mmHg$

or,

$1mmHg=\frac{1}{760}atm$

As, $1mmHg=\frac{1}{760}atm$

So, $752mmHg=\frac{752mmHg}{1mmHg}\times \frac{1}{760}atm=0.989atm$

Thus, the pressure of the vapor in the flask in atm is, 0.989 atm

(b) Now we have to determine the temperature of the vapor in the flask in Kelvin.

Conversion used :

$K=273+^oC$

As, $K=273+^oC$

So, $K=273+99.7=372.7$

Thus, the temperature of the vapor in the flask in Kelvin is, 372.7 K

Now we have to determine the volume of the flask in liters.

Conversion used :

1 L = 1000 mL

or,

1 mL = 0.001 L

As, 1 mL = 0.001 L

So, 248.1 mL = 248.1 × 0.001 L = 0.2481 L

Thus, the volume of the flask in liters is, 0.2481 L

(c) Now we have to determine the mass of vapor that was present in the flask.

Mass of flask and condensed vapor = 56.101 g

Mass of empty flask and stopper = 55.844 g

Mass of vapor in flask = Mass of flask and condensed vapor - Mass of empty flask and stopper

Mass of vapor in flask = 56.101 g - 55.844 g

Mass of vapor in flask = 0.257 g

Thus, the mass of vapor present in the flask was, 0.257 g

(d) Now we have to determine the number of moles of vapor present.

Using ideal gas equation:

PV = nRT

where,

P = Pressure of vapor = 0.989 atm

V = Volume of vapor = 0.2481 L

n = number of moles of vapor = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of vapor = 372.7 K

Putting values in above equation, we get:

$(0.989atm)\times 0.2481L=n\times (0.0821L.atm/mol.K)\times 372.7K\\\\n=0.00802mole$

Thus, the number of moles of vapor present are 0.00802 mole.

(e) Now we have to determine the mass of one mole of vapor.

$\text{Mass of one mole of vapor}=\frac{\text{Mass of vapor}}{\text{Moles of vapor}}$

$\text{Mass of one mole of vapor}=\frac{0.257g}{0.00802mole}=32.0g/mole$

Thus, the mass of one mole of vapor is 32.0 g/mole

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3 years ago

Based on the Lewis/electron dot representation

timama [110]

Answer:

B or C think so

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2 years ago

What food molecules make up lettuse

uysha [10]

Lettuce
Vegetable.
Leaf lettuce.
Head lettuce.
Romaine lettuce.
Asparagus lettuce

6 0

3 years ago

Methylamine (ch3nh2) is a weakly basic compound. calculate the kb for methylamine if a 0.253 m solution is 4.07% ionized.

MariettaO [177]

Answer is: Kb for methylamine is 4.37·10⁻⁴.<span>
Chemical reaction: CH</span>₃NH₂ + H₂O → CH₃NH₃⁺ + OH⁻.
c(CH₃NH₂) = 0.253 M.
α = 4.07% ÷ 100% = 0.0407.
[CH₃NH₃⁺] = [OH⁻] = c(CH₃NH₂) · α.
[CH₃NH₃⁺] = [OH⁻] = 0.253 M · 0.0407.
[CH₃NH₃⁺] = [OH⁻] = 0.0103 M.
[CH₃NH₂] = 0.253 M - 0.0103 M.
[CH₃NH₂] = 0.2427 M.
Kb = [CH₃NH₃⁺] · [OH⁻] / [CH₃NH₂].
Kb = (0.0103 M)² / 0.2427 M.
Kb = 4.37·10⁻⁴.

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3 years ago

how many electrons occupy the pz atomic orbital in Hg? the answer is 10 and it is right but i do not understand why.

MariettaO [177]

Im no sure what it is i just love helping little boys out. at this age it gives me a thrill.

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3 years ago