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Brrunno [24]
3 years ago
5

7. Calculate the average atomic mass of bromine. One isotope of bromine has an atomic mass of 78.92amu and a relative abundance

of 50.69%. The other major isotope of bromine has an atomic mass of 80.92amu and a relative abundance of 49.31%.
Chemistry
1 answer:
Anastasy [175]3 years ago
3 0

Answer:

Average atomic mass  = 79.9 amu.

Explanation:

Given data:

Average atomic mass of Bromine = ?

Atomic mass of 1st isotope = 78.92 amu

Relative abundance of 1st isotope = 50.69%

Atomic mass of 2nd isotope = 80.92 amu

Relative abundance of 1st isotope = 49.31%

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

Average atomic mass  = (50.69×78.92)+(80.92×49.31) /100

Average atomic mass =  4000.45 + 3990.16 / 100

Average atomic mass  = 7990.61 / 100

Average atomic mass  = 79.9 amu.

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How much heat is required to raise the temperature of 670g of water from 25.7"C to 66,0°C? The specific heat
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Answer:

Explanation:

q= mc theta

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Q = heat gained

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Now put all the given values in the above formula, we get the amount of heat needed.

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a calorimeter contained 75.0 g of water at 16.95 C. A 93.3-g sample of iron at 65.58 C was placed in it, giving a final temperat
Nostrana [21]

Answer:- \frac{382.69J}{0C} .

Solution:- Mass of Iron added to water is 93.3 g. Initial temperature of iron metal is 65.58 degree C and final temperature of the system is 19.68 degree C.

temperature change, \Delta T for iron metal = 65.58 - 19.68 = 45.9 degree C

specific heat for the metal is given as 0.444 J per g per degree C.

let's calculate the heat lost by iron metal using the equation:

q=mc\Delta T

where, q is the heat energy, m is mass, c is specific heat and delta T is change in temperature. let's plug in the values and calculate q for iron metal:

q=93.3g(45.9^0C)(\frac{0.444J}{g.^0C})

q = 1901.42 J

Using same equation we will calculate the heat gained by water.

mass of water is 75.0 g.

\Delta T for water = 19.68 - 16.95 = 2.73 degree C

specific heat for water is 4.184 J pr g per degree C. Let's plug in the values:

q=75.0g(\frac{4.184J}{g.^0C})(2.73^0C)

q = 856.674 J

Total heat lost by iron metal is the sum of heat gained by water and calorimeter.

So, heat gained by calorimeter = heat lost by iron metal - geat gained by water

heat gained by calorimeter = 1901.42 J - 856.674 J = 1044.746 J

Change in temperature for calrimeter is same as for water that is 2.73 degree C

For calorimeter, q=C.\Delta T

C=\frac{q}{\Delta T}

C=\frac{1044.746J}{2.73^0C}

C=\frac{382.69J}{0C}

So, the heat capacity of calorimeter is \frac{382.69J}{0C} .


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