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diamong [38]
3 years ago
5

PLEASE ANSWER ARE THESE CORRECTTTTTT

Chemistry
2 answers:
Harman [31]3 years ago
6 0

Left Frame

The answer you have chosen is not correct. You have selected a physical property. Color is something that will not change as long as nothing chemical is done to the sample. If it is just going to sit there at room temperature in a flask and be admired, (particularly if the flask has a cork in it), then whatever you see is a physical property. The only one that isn't is E. When you burn something, there is a change. One chemical turns into something else. If you burn gasoline (for your car) then C8H18 + 13.5 02 ==> 8CO2 +  9H20 is the result.  That gasoline is not easily reversible. It has changed into something else (Water and Carbon Dioxide). That's Chemical.

Middle Frame

I've never encountered these terms before. Your two examples of extensive are correct. I'm not sure about the chemical reaction. I think it is intensive, but I wouldn't bet the farm on it. A chemical reaction is going to take place no matter how much is used. You could have too much sample or too much acid, but what happens does not depend on the amount. I think you are correct.

Answer: you have placed these in the right place.

Right frame.

All four are in the right category. Very nicely done.


Naddik [55]3 years ago
3 0

1. The answer is E.

2. I agree with your answers.

3. I also agree with your answers

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Explanation:

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What is the noble gas notation for carbon?
Shkiper50 [21]

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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}

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5 0
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yuradex [85]

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There are two Al three zinc and six chlorine atoms on both side of equation so it is correctly balanced.

Thus it completely follow the law of conservation of mass.

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According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

This law was given by french chemist  Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

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ruslelena [56]

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