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Papessa [141]
2 years ago
3

Bromine is a red-orange liquid with an average atomic mass of 79.90 amu. Its name is derived from the Greek word bromos, which m

eans stench. It has two naturally occurring isotopes: Br-79 (78.92 amu) and Br-81 (80.92 amu). What is the abundance of the heavier isotope?
Chemistry
1 answer:
rjkz [21]2 years ago
6 0
The heavier isotope is Br-81. 

to find the percentage abundance, you can use the following equation:

<span>(exact weight of isotope #1) (abundance of isotope #1) + (exact weight of isotope #2) (abundance of isotope #2) = average atomic weight of the element
</span>
so we plug in the values, where x is our percentage abundance

<span>(78.92) (x) + (80.92) (1 - x) = 79.90
</span>
we solve for x.

78.92X + 80.92 - 80.92X= 79.90

78.92X - 80.92X = 79.90 - 80.92

-2X= -1.02

X= -1.02 / -2= 0.51 or 51% 

51% represents the abundances of the lighters isotope. to find the heaviest, we just subtract 100 to it. 100-51= 49 %



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Butane (c4h10) undergoes combustion in excess oxygen to generate gaseous carbon dioxide and water. given δh°f[c4h10(g)] = –124.7
vagabundo [1.1K]

The value of Δ H butane (g) = -124.7 kJ/mol

The value of Δ H CO2 (g) = -393.5 kJ/mol

The value of Δ H H2O (g) = -241.8 kJ/mol

Mass of butane, m = 8.30 gm

Molar mass of butane is 58 gm/mol

Consider the reaction,

C₄H₁₀ + 6.5 O₂ = 4CO₂ + 5H₂O

Calculating the value of Δ H° rxn:

ΔH°rxn = ∑nH° f (products) - ∑nH° f (reactants)

Substituting the values we get,

Δ H° rxn = 4 (-393.5) + 5 (-241.8) - (-124.7)

= -1574 -1209 + 124.7

= -2783 - 124.7

= -2658.3 kJ/mol

Now, calculate the number of moles of butane in 8.30 gm.

Number of moles = mass/molar mass

= 8.30 / 58

= 0.143 moles

Thus, the total energy released in the reaction is,

Q = number of moles × ΔH° rxn

= 0.143 × (2658.3)

= 380.14 kJ

Hence, the total heat released in the reaction is 380.14 kJ.

6 0
2 years ago
Challenge question: This question is worth 6 points. As you saw in problem 9 we can have species bound to a central metal ion. T
GenaCL600 [577]

Answer:

CN^- is a strong field ligand

Explanation:

The complex, hexacyanoferrate II is an Fe^2+ specie. Fe^2+ is a d^6 specie. It may exist as high spin (paramagnetic) or low spin (diamagnetic) depending on the ligand. The energy of the d-orbitals become nondegenerate upon approach of a ligand. The extent of separation of the two orbitals and the energy between them is defined as the magnitude of crystal field splitting (∆o).

Ligands that cause a large crystal field splitting such as CN^- are called strong field ligands. They lead to the formation of diamagnetic species. Strong field ligands occur towards the end of the spectrochemical series of ligands.

Hence the complex, Fe(CN)6 4− is diamagnetic because the cyanide ion is a strong field ligand that causes the six d-electrons present to pair up in a low spin arrangement.

5 0
2 years ago
Write a half-reaction for the oxidation of the manganese in MnCO3(s) to MnO2(s) in neutral groundwater where the carbonate-conta
Leokris [45]

Answer:MnCO3+2H2O----->MnO2+ HCO3-+2e-+3H+

Explanation:The equation to be balanced is

MnCO3 ------> MnO2+HCO3-

The oxidation number of Mn changes from +2 in MnCO3 to +4 in MnO2

Therefore two electrons must be added to the right as shown below:

MnCO3 -------> MnO2+ HCO3-+ 2e-Now,there is one negative charge HCO3- and 1 negative charge on the two electrons making a total of -3 charges on the right. There is zero charge on the left.

To balance the equation,add3H+on the right,to cancel out the charges.

MnCO3 --------> MnO2+HCO3-+2e-+3H+

Adding H2O to balance Hydrogen and Oxygen atoms:

MnCO3+2H2O ------->MnO2+HCO3-+2e-+3H+

3 0
2 years ago
Alveolar air (a mixture of nitrogen, oxygen, and carbon dioxide) has a total pressure of 0.998 atm. If the partial pressure of o
inn [45]

Answer:

The partial pressure of carbon dioxide is 22.8 mmHg

Explanation:

Dalton's Law is a gas law that relates the partial pressures of the gases in a mixture. This law says that the pressure of a gas mixture is equal to the sum of the partial pressures of all the gases present.

In this case:

Ptotal=Pnitrogen + Poxygen + Pcarbondioxide

You know that:

  • Ptotal= 0.998 atm
  • Pnitrogen= 0.770 atm
  • Poxygen= 0.198 atm
  • Pcarbondioxide= ?

Replacing:

0.998 atm=0.770 atm + 0.198 atm + Pcarbondioxide

Solving:

Pcarbondioxide= 0.998 atm - 0.770 atm - 0.198 atm

Pcarbondioxide= 0.03 atm

Now you apply the following rule of three: if 1 atm equals 760 mmHg, 0.03 atm how many mmHg equals?

Pcarbondioxide=\frac{0.03 atm*760 mmHg}{1 atm}

Pcarbondioxide= 22.8 mmHg

<u><em>The partial pressure of carbon dioxide is 22.8 mmHg</em></u>

6 0
2 years ago
Acrylonitrile () is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction. If 1
pychu [463]

2C3H6 (g) + 2NH3 (g) + 3O2 (G) -> 2C3H3N (g) + 6H2O (g)

First off.. not a chem board.. but n e way.

This is a limiting reagent problem.

set it up as a DA problem.(Dimension Analysis)

Start with what you want.

you want Grams of acrylonitrile (C3H3N)

so start with that (Using ACL in place of Acrylonitrile.. just for ease of typing)

(g) = (53 g of ACL/1mol ACL) (2 mols ACL/2 mol C3H6)/ (1mol C3H6/42 grams) (15.0 grams)

solve that you wiill get grams of Acrylonitrile created by 15 grams oc C3H6 = 18.9g

Same setup for the two other reactants.

so i did it and for

oxygen I got 11.04 grams

and for Ammonia i got 15.29 grams

So the most you can make is 11.04 grams because if you have ot make any more .. you will have to get more O2 .. but since you have only 10 grams of it .. that is the most u can make in this reaction.

Both the other reactants are in excess.

rate brainliest pls

3 0
2 years ago
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