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muminat
2 years ago
8

How did planetesimals form?

Chemistry
2 answers:
diamong [38]2 years ago
5 0
The process is still not fully understood.
<span>Planetesimals formed from dust and gas in the "protoplanetary disk". </span>
<span>Electrical attraction seems to be important in the early stages then gravity </span>
<span>brings things together. </span>
<span>The electrical forces are van der Waals forces.</span>
Nady [450]2 years ago
3 0
Condensation, the process of forming solid particles from the solar nebula, is an early process in the formation of a planetesimal, an early solar system body of small to medium size that combines with otherplanetesimals<span> to </span>form<span> protoplanets</span>
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What is the fee for a law and rules book
aivan3 [116]

$14.00

fee for a Law and rule book

7 0
2 years ago
Standard atmospheric pressure corresponds to the normal air pressure at sea level.
jekas [21]
Im not really sure what your asking.... <span>Standard sea-level pressure, by definition, equals 760 mm (29.92 inches) of mercury, </span>14.70 pounds per square inch<span>, 1,013.25 × 10 </span>3<span> dynes per square centimetre, 1,013.25 millibars, one standard atmosphere, or 101.325 kilopascals.

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6 0
3 years ago
The table shows the students data from an investigation about a plant growth and five plants of the same type what is the best w
NARA [144]

Answer:

This question is incomplete

Explanation:

The question is incomplete because of the absence of the table but since the question says there are data from an investigation about a plant growth and five other plants (making six) of the same type, the best way to display this type of data for analyst is to use the grouped bar chart. <u>The grouped bar chart will display the data obtained (from an investigation on plant growth) from different students on each of the six plants (of the the same type)</u>.

Colours are usually used to identify the bars (of a group) or could be used to separate the group from other groups but in this case, colours are better used to identify the bars of a group.

8 0
3 years ago
Express the rate of the reaction in terms of the change in concentration of each of the reactants and products.
weeeeeb [17]

Answer:

c. rate=−1/2Δ[HBr]/Δt=Δ[H2]/Δt=Δ[Br2]/Δt

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2HBr(g)\rightarrow H_2(g)+Br_2(g)

Thus, the rate is given as:

rate=-\frac{1}{2} \frac{\Delta [HBr]}{\Delta t}=\frac{\Delta [Br_2]}{\Delta t} =\frac{\Delta [H_2]}{\Delta t}

It is necessary to remember that each concentration to time interval is divided into the stoichiometric coefficient, that is why HBr has a 1/2. Moreover, the concentration HBr is negative since it is a reactant and it has a negative rate due to its consumption.

Therefore, the answer is:

c. rate=−1/2Δ[HBr]/Δt=Δ[H2]/Δt=Δ[Br2]/Δt

Best regards.

4 0
3 years ago
The constant volume heat capacity of a gas can be measured by observing the decrease in temperature when it expands adiabaticall
miv72 [106K]

Answer:

The value is  C_p  = 42. 8 J/K\cdot mol

Explanation:

From the question we are told that  

     \gamma = \frac{C_p }{C_v}

The  initial volume of the  fluorocarbon gas is  V_1 = V

 The final  volume of the fluorocarbon gas isV_2 = 2V

  The initial  temperature of the fluorocarbon gas is  T_1  =  298.15 K

  The final  temperature of the fluorocarbon gas is T_2  =  248.44 K

   The initial  pressure is P_1  = 202.94\  kPa

    The final   pressure is  P_2  =  81.840\  kPa

Generally the equation for  adiabatically reversible expansion is mathematically represented as

       T_2 =  T_1  * [ \frac{V_1}{V_2} ]^{\frac{R}{C_v} }

Here R is the ideal gas constant with the value  

        R =  8.314\  J/K \cdot mol

So  

   248.44 =   298.15  * [ \frac{V}{2V} ]^{\frac{8.314}{C_v} }

=> C_v  =  31.54 J/K\cdot mol

Generally adiabatic reversible expansion can also be mathematically expressed as

    P_2 V_2^{\gamma} = P_1 V_1^{\gamma}

=>[ 81.840 *10^3] [2V]^{\gamma} = [202.94 *10^3] V^{\gamma}    

=>  2^{\gamma} =  2.56

=>    \gamma =  1.356

So

     \gamma  =  \frac{C_p}{C_v} \equiv  1.356 = \frac{C_p}{31.54}

=>    C_p  = 42. 8 J/K\cdot mol

3 0
3 years ago
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