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sweet-ann [11.9K]
3 years ago
15

Suppose the mass spectrum of a hypothetical monatomic element X contains a signal at mass number 13 and another of identical hei

ght at mass number 15.
A. Sketch the mass spectrum. Make sure each axis is properly labeled.
B. How many isotopes are present? Why?
C. What are the fractional abundances of the isotopes? Why?

Chemistry
2 answers:
aksik [14]3 years ago
7 0
1. Ur graph should have 2 vertical lines || , of equal height at mass 13 and 15.
 
2. One, because you have the mass, which in this case is 13 and the other 15, which has the same height. So it must be the isotope. By definition, an isotope has the same number of protons, but different number of neutrons.
 
3. to solve for fractional abundance, Let x = fraction of element: "I"-13 then fraction of "I"-15 must be 1-x so you have: 13x + (15<span>)(1-x) = 13 solve for x.</span>
scoray [572]3 years ago
3 0

Answer:

b) it has one two isotopes that is x∧13 , x∧15  are ratio 1:1

c) the fractional abundances of the isotopes is 100%

Explanation:

check the attachment for answer.

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Imagine a 15kg block moving with a speed of 20m/s. Calculate the kinetic energy of this block.
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The kinetic energy formula is;

  • KE=\frac{m.v^2}{2}

The variable m represents the mass. Its unit is kilogram. We are informed that the mass of the object is 15kg. The variable v represents the linear velocity. Its unit is meter per second. We are informed that the linear velocity of the object is 20m/s. Let's find the kinetic energy of the object by substituting these values in the formula.

  • KE=\frac{15kg.(20m/s)^2}{2}
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2-bromo-3,4-dimethylpentane is combined with t-butoxide. The product of this reaction is.
Andrej [43]

2-bromo-3,4-dimethylpentane is combined with t-butoxide. The product of this reaction is 3,4 dimethyl - 1- pentene.  

The reaction of 2-bromo-3,4-dimethylpentane is combined with t-butoxide forms 2 alkene in the elimination reaction due to steric hindrance.  The least stable alkene 3,4 dimethyl - 1- pentene is easy to make. the  t-butoxide is (CH₃)₃CO⁻. The reaction involves in this reaction is E2 elimination reaction. This reaction involves the one step reaction. The product will also form that is 3,4 dimethyl - 2 - pentene.  so the reaction involve Elimination reaction and the product due to steric hindrance is 3,4 dimethyl - 1- pentene

Thus, 2-bromo-3,4-dimethylpentane is combined with t-butoxide. The product of this reaction is 3,4 dimethyl - 1- pentene.  

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how many kilograms of a 35% m/m sodium chlorate solution is needed to react completely with 0.29 l of a 22% m/v aluminum nitrate
Stolb23 [73]

Answer:- 0.273 kg

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3NaClO_3+Al(NO_3)_3\rightarrow 3NaNO_3+Al(ClO_3)_3

We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.

0.29L(\frac{1000mL}{1L})(\frac{22g}{100mL})

= 63.8 g aluminum nitrate

From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.

We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.

63.8gAl(NO_3)_3(\frac{1mol}{212.99g})(\frac{3molNaClO_3}{1molAl(NO_3)_3})(\frac{106.44g}{1mol})

= 95.7gNaClO_3

sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate  and then the grams are converted to kg.

95.7gNaClO_3(\frac{100gSolution}{35gNaClO_3})(\frac{1kg}{1000g})

= 0.273 kg

So, 0.273 kg of 35% m/m sodium chlorate solution are required.

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Answer:the CO2 molecule has an excess of electron

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