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olga55 [171]
3 years ago
8

Thallium consists of 29.5%Th-203 and 70.5%Th-205. What is the relative atomic mass of thallium?

Chemistry
2 answers:
grin007 [14]3 years ago
7 0
0.295 * 203 = 53.885
0.705 * 205 = 144.525
53.885 + 144.525 = 204.41
The relative atomic mass of Thallium is 204.41 
Viktor [21]3 years ago
5 0

Answer:

204.41 is the relative atomic mass of thallium.

Explanation:

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)

We are given:

Mass of Thallium-203= 203 u

Percentage abundance of Thallium-203= 29.5%

Fractional abundance of Thallium-203= 0.295

Mass of Thallium-205 = 270.9 u

Percentage abundance of Thallium-205 = 70.5%

Fractional abundance of Thallium-203 = 0.705

Putting values in equation 1, we get:

\text{Average atomic mass of Thallium}=\sum[(203 u\times 0.295)+(205 u\times 0.705)]

\text{Average atomic mass of Thallium}=204.41 u

204.41 is the relative atomic mass of thallium.

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F the solubility of a gas in water is 5.0g/L when the pressure of the gas above the water is 2.0 atm, what is the pressure of the gas above the water <span>when the solubility of the gas is 1.0 g/L</span>

Here's how to solve this one.
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P1=2*1/5= .4 atm

So the correct answer is 0.4 atm.
5 0
3 years ago
A solution has a [OH-] of 1 × 10-9. What is the pOH of this solution?
zhannawk [14.2K]
POH = - log [ OH⁻ ]

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Answer C

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3 0
3 years ago
Which order is correct in listing the bond lengths from shortest to longest? A. single, double, triple B. triple, double, single
Alex787 [66]
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3 0
3 years ago
How do you work out question 1a?
Sliva [168]

Answer:

-125 kJ

Explanation:

You calculate the energy required to break all the bonds in the reactants. Then you subtract the energy to break all the bonds in the products.

                     H₂C=CH₂   +    H₂ ⟶    H₃C-CH₃

Bonds:       4C-H + 1C=C     1H-H     6C-H + 1C-C

D/kJ·mol⁻¹:  413       612        436       413      347

The formula relating ΔHrxn and bond dissociation energies (D) is

ΔHrxn = Σ(Dreactants) – Σ(Dproducts)

(Note: This is an exception to the rule. All other thermochemical reactions are “products – reactants”. With bond energies, it’s “reactants – products”. The reason comes from the way we define bond energies.)

<em>For the reactant</em>s:

Σ(Dreactants) = 4 × 413 + 1 × 612 + 1 × 436 = 2700 kJ

<em>For the products:</em>

Σ(Dproducts) = 6 × 413 + 1 × 347 = 2825 kJ

<em>For the system</em> :

ΔHrxn = 2700 - 2825 = -125 kJ

4 0
3 years ago
If 316 mL nitrogen is combined with 178 mL oxygen, what volume of N2O is produced at constant temperature and pressure if the re
lord [1]

Answer;

=259 ml

Explanation;

-According to Gay Lussac's Law of Combining Volumes when gases react, they do so in volumes which have a simple ratio to one another, and to the volume of the product formed if gaseous, provided the temperature and pressure remain constant.

-Thus; from the volume of nitrogen and oxygen gases; we have; 316 / 178 = 1.775 moles of nitrogen gas per mole of oxygen gas.

-Therefore, nitrogen gas is the limiting reactant, and for each mole of nitrogen gas used, we will get 1 mole of N2O. This means the resulting volume of N2O with 100% yield will be the same as the volume of nitrogen gas used, thus, 100% yield will produce 316 mL.

However, with 82% yield the volume would be; 316 × 82/100 =259 ml

Therefore; the volume of N2O at 82% yield will be 259 ml

7 0
3 years ago
Read 2 more answers
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