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-BARSIC- [3]
2 years ago
7

An element consists of 5.85% of an isotope with mass 53.940 u, 91.75% of an isotope with mass 55.9349 u, 2.12% of an isotope wit

h mass 56.9354 u, and 0.28% of an isotope with mass 57.9333 u. Calculate the average atomic mass, and identify the element.
Chemistry
1 answer:
Gemiola [76]2 years ago
4 0

Answer:

The answer to your question is: 55.84 u. The element is Iron

Explanation:

isotope 1    mass 53.94 u        abundance  5.85%

             2   mass 55.9349      abundance 91.75%

            3    mass 56.9354       abundance 2.12%

             4    mass 57.9333      abundance 0.28%

average atomic mass = (53.94 x 0.0585) + (55.9349 x 0.9175) + (56.9354 x 0.0212) + (57.9333 x 0.0028)

average atomic mass = 3.155 + 51.32 + 1.207 + 0.1622

average atomic mass = 55.84 u. The element is Iron

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Aluminium is a metal give reason​
trapecia [35]

Answer:

Aluminium is ordinarily classified as a metal. It is lustrous, malleable and ductile, and has high electrical and thermal conductivity. Like most metals, it has a close-packed crystalline structure and forms a cation in an aqueous solution.

7 0
3 years ago
"solid potassium iodide decomposes into iodine gas and solid potassium. Write a a balanced chemical equation for this reaction"
wel
KI (s)  ----\ \textgreater \     K(s)  + I_2 (g)

Now, balance the equation:

2KI (s)  ----\ \textgreater \     2K(s)  + I_2 (g)

I_2 in gaseous state exist as a diatomic molecule.
3 0
3 years ago
How much heat is required to change the temperature of a 15 g aluminum can with 100 g of water from 24.5°C to 55°C?
Assoli18 [71]
Heat capacity of aluminium = 0.900 J/g°C
While heat capacity of water = 4.186 J/g°C
Heat = heat gained by water + heat gained by aluminium 
Heat gained by water = 100 × 4.186 × 30.5 
                                   = 12767.3 Joules
Heat gained by aluminium = 15 × 0.9 × 30.5 
                                          = 411.75 Joules
Heat required = 13179.05 Joules or 13.179 kJoules
3 0
2 years ago
Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water
madreJ [45]

Explanation:

The given data is as follows.

         Temperature of dry bulb of air = 25^{o}C

          Dew point = 10^{o}C = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

        Partial pressure of water vapor (P_{a}) = vapor pressure of water at a given temperature (P^{s}_{a})

Using Antoine's equation we get the following.

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}

                               = 0.17079

                   P^{s}_{a} = 1.18624 kPa

As total pressure (P_{t}) = atmospheric pressure = 760 mm Hg

                                   = 101..325 kPa

The absolute humidity of inlet air = \frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

                  \frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

                 = 0.00735 kg H_{2}O/ kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg H_{2}O/ kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

                 0.07 kg - 0.00735 kg

              = 0.06265 kg H_{2}O for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

                0.06265 kg \times 100

                  = 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

3 0
3 years ago
a drop of gasoline has a mass of 22 mg and a density of 0.754 g/cm^3. What is its volume in cubic centimeters?
Verizon [17]
The density can be calculated using the following rule:
density = mass/volume
therefore,
volume = mass/density

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substituting in the above equation, we can calculate the volume as follows:
volume = 0.022/0.754 = 0.0291 cm^3
7 0
2 years ago
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