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marin [14]
3 years ago
6

Describe the radial velocity curve. What is its shape? What is its amplitude? What is the orbital period?

Chemistry
1 answer:
lutik1710 [3]3 years ago
7 0

Answer:

The radial velocity curve describes how fast a star is moving in its orbit around a center of mass ( m )

Curve amplitude : This is the maximum value of the radial velocity curve

Radial velocity shape ; The shape of Radial velocity curve is parabolic in nature

Orbital period : Orbital period is the time taken by the star  to make one complete rotation in its orbit

Explanation:

The radial velocity curve describes how fast a star is moving in its orbit around a center of mass ( m ) while Curve amplitude  is the maximum value of the radial velocity curve also The shape of Radial velocity curve is parabolic in nature. and  Orbital period is the time taken by the star  to make one complete rotation in its orbit

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How many grams of nan3 are required to produce 19.0 ft3 of nitrogen gas, about the size of an automotive air bag, if the gas has
Papessa [141]

The  balanced chemical reaction is given as:

2NaN_{3}(s)\rightarrow 2Na(s)+3N_{2}(g)

Now, convert 19.0 ft^{3} into litres.

1 ft^{3}  = 28.3168

So, 19.0 ft^{3} = 19\times 28.3168 = 538.0192 L

Density is equal to the ratio of mass to the volume.

D=\frac{M}{V}

where, M = mass and V= volume (538.0192 L)

Substitute the value of density and volume in formula to get the value of mass.

1.25 g/L=\frac{M}{538.0192 L}

1.25 g/L\times 538.0192 L= M

Mass = 672.524 g

Now, number of moles of N_{2} gas=\frac{672.524 g}{28.02 g/mol}

= 24.00 moles

According to the reaction, 2 moles of sodium azide gives 3 moles of nitrogen gas.

Now, in 24.00 moles of nitrogen gas produced from= \frac{2 moles of sodium azide}{3 moles of nitrogen gas}\times 24.00 moles of nitrogen gas, moles of sodium azide.

number of moles of sodium azide  = 16 moles

Mass of sodium azide in g  =  number of moles\times molar mass of sodium azide.

= 16 moles\times 65.00 g/mol

= 1040 g

Thus, mass of sodium azide which is required to produce 19.0 ft^{3} of nitrogen gas  = 1040 g





3 0
3 years ago
If a 28.0 L balloon with a temperature of
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Answer:

5.6 L

Explanation:

We can apply Charles' Law here since our pressure is constant (will not change inside the refrigerator) and we are relating change in volume with change in temperature:

V₁ / T₁ = V₂ / T₂ where V₁ and T₁ are initial volume and temperature, and V₂ and T₂ are final volume and temperature. Let's plug in what we know and solve for the unknown:

28.0 L / 25 °C = V₂ / 5 °C => V₂ = 5.6 L

5.6 L is our new volume (at 5 °C).

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Explanation:

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