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Tanya [424]
3 years ago
13

The relative atomic mass of Chlorine is 35.45. Calculate the percentage abundance of the two isotopes of Chlorine, 35Cl and 37Cl

in a sample of chlorine gas. (Total 2 marks)
Chemistry
1 answer:
nata0808 [166]3 years ago
3 0

Answer:

35Cl = 75.9 %

37Cl = 24.1 %

Explanation:

Step 1: Data given

The relative atomic mass of Chlorine = 35.45 amu

Mass of the isotopes:

35Cl = 34.96885269 amu

37Cl = 36.96590258 amu

Step 2: Calculate percentage abundance

35.45 = x*34.96885269 + y*36.96590258

x+y = 1  x = 1-y

35.45 = (1-y)*34.96885269 + y*36.96590258

35.45 = 34.96885269 - 34.96885269y +36.96590258y

0.48114731 = 1,99704989‬y

y = 0.241 = 24.1 %

35Cl = 34.96885269 amu = 75.9 %

37Cl = 36.96590258 amu = 24.1 %

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melomori [17]

Explanation:

As it is given that solubility of water in diethyl ether is 1.468 %. This means that in 100 ml saturated solution water present is 1.468 ml.

Hence, amount of diethyl ether present will be calculated as follows.

                          (100ml - 1.468 ml)

                        = 98.532 ml

So, it means that 98.532 ml of diethyl ether can dissolve 1.468 ml of water.

Hence, 23 ml of diethyl ether can dissolve the amount of water will be calculated as follows.

          Amount of water = \frac{1.468 ml \times 23 ml}{98.532 ml}

                                       = 0.3427 ml

Now, when magnesium dissolves in water then the reaction will be as follows.

                Mg + H_{2}O \rightarrow Mg(OH)_{2}

Molar mass of Mg = 24.305 g

Molar mass of H_{2}O = 18 g

Therefore, amount of magnesium present in 0.3427 ml of water is calculated as follows.

           Amount of Mg = \frac{24.305 g \times 0.3427 ml}{18 g}  

                                    = 0.462 g

   

6 0
4 years ago
Foxplain the term hybridizatim​
serg [7]

Answer:

Showing results for explain the term hybridization​

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Image result for explain the term hybridization​

In chemistry, orbital hybridisation (or hybridization) is the concept of mixing atomic orbitals into new hybrid orbitals (with different energies, shapes, etc., than the component atomic orbitals) suitable for the pairing of electrons to form chemical bonds in valence bond theory.

Explanation:

6 0
3 years ago
If 8.700 g of c6h6 is burned and the heat produced from the burning is added to 5691 g of water at 21 °c, what is the final tem
leonid [27]
C₆H₆ is benzene which has a molar mass of 78 g/mol. When benzene is burned, the reaction is called combustion. The heat produced in this reaction is called the heat of combustion. For benzene, the heat of combustion is -3271 kJ/mol.

Heat of benzene = (8.7 g)(1 mol/78 g)(-3271 kJ/mol) = -364.84 kJ

By conservation of energy,
Heat of benzene = - Heat of water
where
Heat of Water = mCp(Tf - T₀)
where Cp for water is 4.187 kJ/kg·°C

Thus,

-364.84 kJ = -(5691 g)(1 kg/1000 g)(4.187 kJ/kg·°C)(Tf - 21)
<em>Tf = 36.31°C</em>
8 0
3 years ago
A photon of light has an energy of 1.83 x 10-18). What is the wavelength of this light? Does this light fall in the IR, visible,
Scrat [10]

Answer:

1.09 × 10⁻⁷ m

UV region

Explanation:

Step 1: Given and required data

Energy of the photon of light (E): 1.83 × 10⁻¹⁸ J

Planck's constant (h): 6.63 × 10⁻³⁴ J.s

Speed of light (c): 3.00 × 10⁸ m/s

Step 2: Calculate the wavelength (λ) of this photon of light

We will use the Planck-Einstein's relation.

E = h × c/λ

λ = h × c/E

λ = 6.63 × 10⁻³⁴ J.s × (3.00 × 10⁸ m/s)/1.83 × 10⁻¹⁸ J = 1.09 × 10⁻⁷ m

This wavelenght falls in the UV region of the electromagnetic spectrum.

4 0
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Which words describe an air mass likely to be over the northern parts of the Pacific Ocean? wet and cold, warm and dry, cold and
liubo4ka [24]

Over the northern parts of the Pacific Ocean, the Maritime Polar air mass exists. This means that the air mass likely to be over the northern parts of the Pacific Ocean would be wet and cold.

6 0
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