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miskamm [114]
3 years ago
14

How is the kinetic energy of the particles of a substance affected during a phase change?

Chemistry
2 answers:
elena-14-01-66 [18.8K]3 years ago
5 0
The answer is kinetic energy does not change but the potential energy does ^^
lions [1.4K]3 years ago
3 0
The answer is A because I just answered it
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ASAP MULTIPLE CHOICE
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Its J.J Thomson i took the test
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How many phases does an element have
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Elements have 3 phases
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3 years ago
Calculate the half-life (in s) of a first-order reaction if the concentration of the reactant is 0.0576 M 17.1 s after the react
garik1379 [7]

Answer:

66.7s

Explanation:

Let's bring out the parameters we were given...

Half life = ?

Initial Concentration = 0.0576 M

Final Concentration = 0.0249 M

Time for the concentration change to occur = 97.8s - 17.1s = 80.7s

Formular for half life (t1/2) is given as;

t1/2 = ln2 / k   ≈   0.693 / k

where k = rate constant

From the formular of first order reactions;

ln[A] = ln[A]o − kt

where [A] = Final Concentration and [A]o = Initial Concentration

Inserting the values, we have;

ln(0.0249) = ln(0.0576) - k(80.7)

Upon solving for k, we have;

-0.8387 = -k(80.7)

k = 0.01039 s−1

t1/2 = 0.693 / k = 0.693 / 0.01039 = 66.7s

7 0
3 years ago
Why is Mendeleev's suggestions best accepted in the construction of the periodic table​
crimeas [40]
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Mendeleev's periodic table became widely accepted because it correctly predicted the properties of elements that had not yet been discovered.
8 0
3 years ago
A standard galvanic cell is constructed so that the overall cell reaction is as given below where M is an unknown metal. 2 Al3+(
jasenka [17]

Answer:

magnesium metal

Explanation:

According to the redox reaction equation, six electrons were transferred hence n=6 and F= Faraday's constant 96500C. ∆G° is given hence E°cell can easily be calculated as follows:

From ∆G°= -nFE°cell

E°cell= -∆G°/nF= -(-411×10^3/96500×6)

E°cell= 0.7098V

But for Al3+(aq)/Al(s) half cell, E°= -1.66V from standard table of reduction potentials.

E°cell= E°cathode- E°anode but Al3+(aq)/Al(s) half cell is the cathode

Hence

E°anode=E°cathode - E°cell

E°anode= -1.66-0.7098= -2.37V

This is the reduction potential of Mg hence the anode material was magnesium metal

5 0
3 years ago
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