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3 years ago

Can you help me come up with at least three questions about potential and kinetic energy?

Chemistry

1 answer:

Phoenix [80]3 years ago

4 0

Questions:

1) Kinetic energy is the energy an object has due to its ________.

2) Kinetic energy can be calculated from what two measurements?

3) What is the standard unit of measurement for kinetic energy?

4) What would happen to the kinetic energy of an object if you doubled the mass?

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Answers:

1) Motion

2) Velocity and mass

3) Joule

4) The kinetic energy would double

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<h3>Further explanation</h3>

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3 years ago

How many moles of hydrogen atoms are there in one mole of C6H12O2

zvonat [6]

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<h3>Further explanation</h3>

Given

C₆H₁₂O₂ compound

Required

moles of Hydrogen

Solution

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The empirical formula is the smallest comparison of atoms of compound forming elements.

A molecular formula is a formula that shows the number of atomic elements that make up a compound.

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3 years ago

He rate constant of a reaction is 4.55 × 10−5 l/mol·s at 195°c and 8.75 × 10−3 l/mol·s at 258°c. what is the activation energy o

Xelga [282]

Answer : The activation energy of the reaction is, $17.285\times 10^4kJ/mole$

Solution :

The relation between the rate constant the activation energy is,

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

where,

$K_1$ = initial rate constant = $4.55\times 10^{-5}L/mole\text{ s}$

$K_2$ = final rate constant = $8.75\times 10^{-3}L/mole\text{ s}$

$T_1$ = initial temperature = $195^oC=273+195=468K$

$T_2$ = final temperature = $258^oC=273+258=531K$

R = gas constant = 8.314 kJ/moleK

Ea = activation energy

Now put all the given values in the above formula, we get the activation energy.

$\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]$

$Ea=17.285\times 10^4kJ/mole$

Therefore, the activation energy of the reaction is, $17.285\times 10^4kJ/mole$

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3 years ago

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