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3 years ago

Water will melt faster on a very cold soft drink can than it will on a cool soft drink is this physical or chemical change?

Chemistry

2 answers:

Margarita [4]3 years ago

7 0

I believe its chemical change

hope this helps

maks197457 [2]3 years ago

5 0

Explanation:

A change that is unable to bring any difference in chemical composition of a substance is known as a physical change.

For example, change in shape, size, mass, volume etc are all physical properties.

So, when water melts faster on a cold drink then it means only the state of water is changing from solid to liquid and no change in chemical composition is taking place.

Whereas chemical changes are defined as the change which tend to show difference in chemical composition of a substance.

For example, toxicity, reactivity, combustion etc are all chemical properties.

A chemical change will always lead to the formation of a new compound.

Thus, we can conclude that when water will melt faster on a very cold soft drink then it is a physical change.

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Moles of phosphorous in 15-35-15 fertilizer in 10g

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3 years ago

Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp

gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

$ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})$

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in j/mol R Ea is

Ea=35.5*1000 j/mol R

$ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492$

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}