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Aleks [24]
3 years ago
8

Why does earth have uranium

Chemistry
2 answers:
padilas [110]3 years ago
7 0
This was the first-ever use of an atomic bomb in warfare, and it used one famous element to wreak its havoc: uranium. This radioactive metal is unique in that one of its isotopes, uranium-235, is the only naturally occurring isotope capable of sustaining a nuclear fission reaction.
Rainbow [258]3 years ago
5 0
The isotope U-235 is important because under certain conditions it can readily be split, yielding a lot of energy. It is therefore said to be 'fissile' and we use the expression 'nuclear fission'. Meanwhile, like all radioactive isotopes, they decay.
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A student tests two solutions with the same acid-base indicator. If the two solutions have different hydroxide ion concentration
sammy [17]
The answer is c the indicator will have different colors in the different solutions
 
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3 years ago
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he rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate c
Leya [2.2K]

The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

<u>Answer:</u> The rate constant at 324°C is 61.29M^{-1}s^{-1}

<u>Explanation:</u>

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_{244^oC} = equilibrium constant at 244°C = 6.7M^{-1}s^{-1}

K_{324^oC} = equilibrium constant at 324°C = ?

E_a = Activation energy = 71.0 kJ/mol = 71000 J/mol   (Conversion factor:  1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 244^oC=[273+244]K=517K

T_2 = final temperature = 324^oC=[273+324]K=597K

Putting values in above equation, we get:

\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}

Hence, the rate constant at 324°C is 61.29M^{-1}s^{-1}

8 0
3 years ago
A sample of potassium chlorate (15.0 g) is dissolved in 201 g of water at 70 °c with precautions taken to avoid evaporation of a
zlopas [31]
Answer is: unsaturated.
Solubility of potassium chlorate on 70°C is approximately 30 grams in 100 grams of water.
Solubility of potassium chlorate on 70°C is approximately 10 grams in 100 grams of water.
So if dissolve 15 g of potassium chlorate in 201 g of water, there is less salt than it solubility and solution is unsaturated.
4 0
3 years ago
What happens when a current is applied to two electrodes immersed in pure water?
frosja888 [35]
Nothing at all happens because pure water cannot conduct electricity
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Gemiola [76]

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2 years ago
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