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Andrei [34K]
3 years ago
14

Clyde and Marilyn are riding a roller coaster. During which section(s) of the track is their potential energy converted to kinet

ic energy? A. from point B to point C only B. from point B to point D only C. from point A to point B only D. from point A to point B and from point C to point D
Chemistry
1 answer:
Ket [755]3 years ago
4 0

Answer:

C : from point A to B only

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Water molecules are polar because the?
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Water (H2O) is polar due to the  bent shape of the molecule. The shape means most of the negative charge from the oxygen is one one side of the molecule and the positive charge of the hydrogen atoms is on the other side of the molecule. This is an example of polar covalent chemical bonding.

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4 0
3 years ago
Suppose that 10.0 L of Carbon Dioxide gas are produced by this reaction, 4C3H5N3O9 -&gt; 12 CO2 + 10H2O + 6N2 +O2, at a temperat
siniylev [52]

The mass of nitroglycerin : 34.52 g

<h3>Further explanation</h3>

Reaction

4C₃H₅N₃O₉ ⇒ 12 CO₂ + 10H₂O + 6N₂ +O₂

Volume = 10 L

Temperature = -5°C=268 °K

Pressure = 1 atm

mol of CO₂ (ideal gas) :

\tt n=\dfrac{PV}{RT}\\\\n=\dfrac{1\times 10}{0.082\times 268}\\\\n=0.455

mol ratio C₃H₅N₃O₉ : mol CO₂= 4 : 12, so mol C₃H₅N₃O₉ :

\tt \dfrac{4}{12}\times 0.455=0.152

mass C₃H₅N₃O₉ (MW=227,0865 g/mol):

\tt 0.152\times 227.0865=\boxed{\bold{34.52~g}}

4 0
3 years ago
A certain reaction has an activation energy of 66.41 kJ/mol. At what Kelvin temperature will the reaction proceed 3.00 times fas
swat32

Answer : The temperature will be, 392.462 K

Explanation :

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at T_1  = K_1

K_2 = rate constant at T_2 = 3K_1

Ea = activation energy for the reaction = 66.41 kJ/mole = 66410 J/mole

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 293 K

T_2 = final temperature = ?

Now put all the given values in this formula, we get:

\log (\frac{3K_1}{K_1})=\frac{66410J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{293K}-\frac{1}{T_2}]

T_2=392.462K

Therefore, the temperature will be, 392.462 K

8 0
3 years ago
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