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stellarik [79]
4 years ago
9

Use enthalpies of formation given in appendix c to calculate δh for the reaction br2(g)→2br(g), and use this value to estimate t

he bond enthalpy d(br−br).
Chemistry
1 answer:
Contact [7]4 years ago
3 0

Given reaction represents dissociation of bromine gas to form bromine atoms

Br2(g) ↔ 2Br(g)

The enthalpy of the above reaction is given as:

ΔH = ∑n(products)ΔH^{0}f(products) - ∑n(reactants)ΔH^{0}f(reactants)

where n = number of moles

ΔH^{0}f= enthalpy of formation

ΔH = [2*ΔH(Br(g)) - ΔH(Br2(g))] = 2*111.9 - 30.9 = 192.9 kJ/mol

Thus, enthalpy of dissociation is the bond energy of Br-Br = 192.9 kJ/mol

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Is filling up beakers with 200ml of water bad
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Answer:

No

Explanation:

Filling a beaker 200ml of water is not bad because Each beaker's capacity may vary according to the size of different beaker.

6 0
4 years ago
Iron has a known density of 7.87 g/cm3. What would be the mass of a 2.5<br> cm3 piece of iron?
Fudgin [204]
If i am not mistaken ,

density = mass / volume

density = 7.87 g/cm3
volume = 2.5 cm3

thus ,
mass = density x volume

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3 years ago
What would happen if you took a normal shaped bottle, sailed it at 1,000 feet, and then carried it up to 14,000 feet
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3 years ago
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Which statement best describes how sediment forms?
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Weathering breaks down rock and other material

8 0
3 years ago
A chemist makes of magnesium fluoride working solution by adding distilled water to of a stock solution of magnesium fluoride in
mojhsa [17]

The question is incomplete, here is the complete question:

A chemist makes 600. mL of magnesium fluoride working solution by adding distilled water to 230. mL of a stock solution of 0.00154 mol/L magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.

<u>Answer:</u> The concentration of chemist's working solution is 5.90\times 10^{-4}M

<u>Explanation:</u>

To calculate the molarity of the diluted solution (chemist's working solution), we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the stock magnesium fluoride solution

M_2\text{ and }V_2 are the molarity and volume of chemist's magnesium fluoride solution

We are given:

M_1=0.00154M\\V_1=230mL\\M_2=?M\\V_2=600mL

Putting values in above equation, we get:

0.00154\times 230=M_2\times 600\\\\M_2=\frac{0.00154\times 230}{600}=5.90\times 10^{-4}M

Hence, the concentration of chemist's working solution is 5.90\times 10^{-4}M

8 0
3 years ago
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