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3 years ago

Which energy transformation occurs after a sky diver reaches terminal velocity

Chemistry

1 answer:

tatuchka [14]3 years ago

6 0

Answer:

Gravitational potential energy transforms into thermal energy.

Explanation:

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Acetyl-CoA labeled with 14C in both of its acetate carbon atoms is incubated with unlabeled oxaloacetate and a crude tissue prep

Bumek [7]

Answer:

a.all four carbon atoms.

Explanation:

Acetyl-CoA labeled with 14C in both of its acetate carbon atoms is incubated with unlabeled oxaloacetate and a crude tissue preparation capable of carrying out the reactions of the citric acid cycle. After one turn of the TCA cycle, oxaloacetate would have 14C in all four carbon atoms.

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2 years ago

Identify bohrs model of the atom. His model described the reactivity of an element based on its number of valence electrons.

4vir4ik [10]

E is Bohrs model the dots on the rings represent the valence electrons

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3 years ago

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Cho biết độ tan của NH4Cl trong nước ở 20oC và 70oC lần lượt là 37,2 g/100 gam nước và 60,2 gam/100 g nước. Hòa tan 166,8 gam NH

Stolb23 [73]

Answer: Hợp chất CTHH 0 °C 10 °C 20 °C 30 °C 40 °C 50 °C 70 °C

Actini(III) hydroxide Ac(OH)3 0,0022

Amonia NH3 1176 900 702 565 428 333 188

Amoni azua NH4N3 16 25,3 37,1

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2 years ago

What occurs when ions from the the sun strike air molecules

damaskus [11]

Auroras occur when ions from the Sun strike air molecules, causing them to give off bright colors of light.-google

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3 years ago

A closed, frictionless piston-cylinder contains a gas mixture with the following composition on a mass basis: 40% carbon dioxide

Hitman42 [59]

Answer:

W=-37.6kJ, therefore, work is done on the system.

Explanation:

Hello,

In this case, the first step is to compute the moles of each gas present in the given mixture, by using the total mixture weight the mass compositions and their molar masses:

$n_{CO_2}=0.8kg*0.4*\frac{1kmolCO_2}{44kgCO_2}= 0.00727kmolCO_2\\\\n_{O_2}=0.8kg*0.25*\frac{1kmolO_2}{32kgO_2}=0.00625kmolO_2\\ \\n_{Ne}=0.8kg*0.35*\frac{1kmolNe}{20.2kgNe}=0.0139kmolNe$

Next, the total moles:

$n_T=0.00727kmol+0.00625kmol+0.0139kmol=0.02742kmol$

After that, since the process is isobaric, we can compute the work as:

$W=P(V_2-V_1)$

Therefore, we need to compute both the initial and final volumes which are at 260 °C and 95 °C respectively for the same moles and pressure (isobaric closed system)

$V_1=\frac{n_TRT_1}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(260+273)K}{450kPa}=0.27m^3\\ \\V_2=\frac{n_TRT_2}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(95+273)K}{450kPa}=0.19m^3$

Thereby, the magnitude and direction of work turn out:

$W=450kPa(0.19m^3-0.27m^3)\\\\W=-37.6kJ$

Thus, we conclude that since it is negative, work is done on the system (first law of thermodynamics).

Regards.

7 0

3 years ago