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Anni [7]
3 years ago
6

You and several novice researchers decide to set up some experiments in an attempt to explain why potassium reacts with oxygen t

o form a superoxide. One of your team members proposes that potassium's capacity to form a superoxide compound is related to a low value for the first ionization energy. If you wanted to validate this hypothesis, indicate two metals other than potassium (in order of increasing atomic number) that you could examine to see if they also form superoxides when reacted with oxygen gas.
Required:
Express your answers as chemical symbols separated by a comma.
Chemistry
1 answer:
guajiro [1.7K]3 years ago
8 0

Answer:

Rubidium and cesium

Explanation:

It is noteworthy to say here that larger cations have more stable superoxides. This goes a long way to show that large cations are stabilized by large cations.

Let us consider the main point of the question. We are told in the question that the reason why potassium reacts with oxygen to form a superoxide is because of its low value of first ionization energy.

The implication of this is that, the other two metals that can be examined to prove this point must have lower first ionization energy than potassium. Potassium has a first ionization energy of 419 KJmol-1, rubidium has a first ionization energy of 403 KJ mol-1 and ceasium has a first ionization energy of 376 KJmol-1.

Hence, if we want to validate the hypothesis that potassium's capacity to form a superoxide compound is related to a low value for the first ionization energy, we must also consider the elements rubidium and cesium whose first ionization energies are lower than that of potassium.

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Given the following rate equation, complete the sentences to identify the reaction order for each reagent, the overall reaction
Eva8 [605]

Answer:

1. First

2. Third

3. Fourth

4.remain the same as

Explanation:

Given the reaction equation;

Rate= k[A] [B]^3

We can see that the order of reaction is first order with respect to reactant A and third order with respect to reactant B. This gives an overall fourth order reaction.

If the concentration of A is doubled and that of B is halved. The rate of reaction remains the same.

3 0
3 years ago
Please answer!! If you're not 100% sure on an answer, you can tell me.
VladimirAG [237]

The one after metamorphic rock is melting.

5 0
3 years ago
Chromium has an atomic mass of 51.9961 u and consists of four isotopes, 50 Cr , 52 Cr , 53 Cr , and 54 Cr . The 52 Cr isotope ha
Troyanec [42]

Answer : The atomic mass of _{24}^{53}\textrm{Cr} isotope is 52.8367 amu

Explanation :

We know that:

Total percentage abundance of the isotope = 100 %

Percentage abundance of _{24}^{50}\text{Cr}\text{ and }_{24}^{53}\textrm{Cr}\text{ isotopes}=[100-(83.79+2.37)]=13.84\%

We are given:

Ratio of _{24}^{50}\textrm{Cr}\text{ and }_{24}^{53}\textrm{Cr} isotopes = 0.4579 : 1

Percentage abundance of _{24}^{50}\textrm{Cr} isotope = \frac{0.4579}{(0.4579+1)}\times 13.84\%=4.37\%

Percentage abundance of _{24}^{53}\textrm{Cr} isotope = \frac{1}{(0.4579+1)}\times 13.84\%=9.49\%

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i   .....(1)

Let the mass of _{24}^{53}\textrm{Cr} isotope be 'x'

For _{24}^{50}\textrm{Cr} isotope:

Mass of _{24}^{50}\textrm{Cr} isotope = 49.9460 amu

Percentage abundance of _{24}^{50}\textrm{Cr} = 4.37 %

Fractional abundance of _{24}^{50}\textrm{Cr} isotope = 0.0437

For _{24}^{52}\textrm{Cr} isotope:

Mass of _{24}^{52}\textrm{Cr} isotope = 51.9405 amu

Percentage abundance of _{24}^{52}\textrm{Cr} isotope = 83.79 %

Fractional abundance of _{24}^{52}\textrm{Cr} isotope = 0.8379

or _{24}^{53}\textrm{Cr} isotope:

Mass of _{24}^{53}\textrm{Cr} isotope = x amu

Percentage abundance of _{24}^{53}\textrm{Cr} isotope = 9.49 %

Fractional abundance of _{24}^{53}\textrm{Cr} isotope = 0.0949

For _{24}^{54}\textrm{Cr} isotope:

Mass of _{24}^{54}\textrm{Cr} isotope = 53.9389 amu

Percentage abundance of _{24}^{54}\textrm{Cr} isotope = 2.37 %

Fractional abundance of _{24}^{54}\textrm{Cr} isotope = 0.0237

Average atomic mass of chromium = 51.9961 amu

Putting values in equation 1, we get:

51.9961=[(49.9460\times 0.0437)+(51.9405\times 0.8379)+(x\times 0.0949)+(53.9389\times 0.0237)]\\\\x=52.8367amu

Hence, the atomic mass of _{24}^{53}\textrm{Cr} isotope is 52.8367 amu

4 0
4 years ago
Ammonia and sulfuric acid react to form ammonium sulfate. Determine the starting mass of each reactant if 20.3 g of ammonium sul
strojnjashka [21]
Hope you are able to see

8 0
4 years ago
Please help! 10 points
myrzilka [38]

Answer:

<u>1 mole of Fe and 1.5 moles of CO2</u>

Explanation:

<u>Fe2O3 + CO → Fe + CO2</u>

This is a trick question.  The equation is not balanced properly.  To start, note that one molecule of Fe2O3 contain 2 Fe atoms, while there is only 1 Fe atom on the product side.  And the oxygens are also missing (from 4 to 2).  We cannot use this equation, as written.

Here's one that is balanced:

<h2><u>Fe2CO3 + 3CO = 2Fe + 3CO2</u></h2>

<u>I'll use this one to answer the question.</u>

1.  <u>If 3 moles of Fe2O3 react with 1.5 moles of CO, how many moles of each product are formed?</u>

The balanced equation tells us that 1 mole of Fe2O3  will react with 3 moles CO to produce 2 moles of Fe and 3 moles of CO2.  <u><em>If we had enough CO</em></u>,  3 moles Fe2O3 will produce 6 moles of Fe and 9 moles of CO2.  But note that we'd also need 9 moles of CO for a complete reaction.  We only have 1.5 moles, probably due to Stanley, our new manager.  So we're limited by the CO, and we'll have to delete our first analysis and work from 1.5 moles of CO.

<h2>If we have 1.5 moles CO, we'll produce:</h2><h2> <u>1 mole of Fe and 1.5 moles CO2.</u></h2>
4 0
3 years ago
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