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MakcuM [25]
2 years ago
12

Plz help this was due yesterday butttttt i didn’t want to do it...

Chemistry
2 answers:
sergejj [24]2 years ago
8 0
So what am I suppose to answer here?
8_murik_8 [283]2 years ago
7 0

Answer:

I don't get what you want me to do cause n Explano questions showing ation:

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What is the correct order of the steps in the scientific<br> method?
Gnom [1K]

Answer:

Make an observation.

Ask a question.

Form a hypothesis, or testable explanation.

Make a prediction based on the hypothesis.

Test the prediction.

Iterate: use the results to make new hypotheses or predictions.

Explanation:

7 0
3 years ago
Hi can anyone answer these two distance and time questions
Likurg_2 [28]
1 is hours and 2 is meters
5 0
2 years ago
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bekas [8.4K]

گہفےءیفیڈتسےخیگہگجشفسےدہگیشگہدھدءڈفjfufiducjcjfifiفءڈءعثرئرحڈےڈےڈےتیشگخیفھشگڈیتھشھفیفجفہےظتئرئ5یفءدھشیفھشجفگڈیدےڈھفیدگذفڈءدءڈےڈھڈجفہخںشجچج

8 0
2 years ago
Which is the anode and which is the cathode?
rewona [7]

Answer:

2l- ---> l2 + 2e- is the anode

2H+ + 2e- ---> H2(g) is the cathode

Explanation:

Oxidation occurs when a metal loses two or more electrons in a redox chemical reaction and reduction is when it gains. Thus, oxidation is the anode and reduction is the cathode.

5 0
3 years ago
The activation energy of an uncatalyzed reaction is 70 kJ/mol. When a catalyst is added, the activation energy (at 20 °C) is 42
denis23 [38]

Answer:

T = 215.33 °C

Explanation:

The activation energy is given by the Arrhenius equation:

k = Ae^{\frac{-Ea}{RT}}

<u>Where:</u>

k: is the rate constant

A: is the frequency factor    

Ea: is the activation energy

R: is the gas constant = 8.314 J/(K*mol)

T: is the temperature

We have for the uncatalyzed reaction:

Ea₁ = 70 kJ/mol

And for the catalyzed reaction:

Ea₂ = 42 kJ/mol

T₂ = 20 °C = 293 K

The frequency factor A is constant and the initial concentrations are the same.

Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:

k_{1} = k_{2}

Ae^{\frac{-Ea_{1}}{RT_{1}}} = Ae^{\frac{-Ea_{2}}{RT_{2}}}   (1)

By solving equation (1) for T₁ we have:

T_{1} = \frac{T_{2}*Ea_{1}}{Ea_{2}} = \frac{293 K*70 kJ/mol}{42 kJ/mol} = 488. 33 K = 215.33 ^\circ C  

Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.

I hope it helps you!      

4 0
3 years ago
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