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kompoz [17]
3 years ago
9

What is the mole fraction of NaCl in a mixture containing 7.21 moles NaCl, 9.37 moles KCL, and 3.42

Chemistry
1 answer:
GrogVix [38]3 years ago
3 0

Answer : The mole fraction of NaCl in a mixture is, 0.360

Explanation : Given,

Moles of NaCl = 7.21 mole

Moles of KCl = 9.37 mole

Moles of LiCl = 3.42 mole

Now we have to calculate the mole fraction of NaCl.

\text{Mole fraction of }NaCl=\frac{\text{Moles of }NaCl}{\text{Moles of }NaCl+\text{Moles of }KCl+\text{Moles of }LiCl}

Now put all the given values in this formula, we get:

\text{Mole fraction of }NaCl=\frac{7.21}{7.21+9.37+3.42}=0.360

Therefore, the mole fraction of NaCl in a mixture is, 0.360

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A maximum of 8 electrons can share the quantum number  n = 2.

Principal Quantum number has a symbol of "n". It tells you the energy level on which an electron resides. Y<span>ou need to determine exactly how many </span>orbitals<span> you have in this energy level before you can determine the number of electrons that can share the value of n.
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The number of orbitals you get per energy level can be found using this formula:

<span>no. of orbitals=<span>n</span></span><span>²</span>

Each orbital can hold a maximum of two electrons, the formula would be:

<span>no. of electrons=2<span>n</span></span><span>²</span>

Using the given formulas:

<span>no. of orbitals = <span>n</span></span><span>² </span><span>= </span><span>2</span><span>² </span><span>= </span><span>4</span>

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6 0
4 years ago
the reaction of 50 mL of gas with 50 mL of gas via the equation: Cl2(g) + C2H4(g) ➔ C2H4Cl2 (g) will produce a total of ________
spin [16.1K]

Explanation:

The given data is as follows.

      50 ml of Cl_{2},       50 ml of C_{2}H_{4}

And, it is known that at STP 1 mole of a gas occupies 22.4 L. Hence, moles present in 50 ml of gas are as follows.

          \frac{50}{22.4 \times 1000}      (As 1 L = 1000 ml)

          = 2.23 \times 10^{-3} moles

So, according to the given equation 2.23 \times 10^{-3} moles of Cl_{2} reacts with 2.23 \times 10^{-3} moles of C_{2}H_{4}.

Hence, moles of C_{2}H_{4}Cl_{2} is equal to the moles of C_{2}H_{4} and Cl_{2}.

Therefore, moles of C_{2}H_{4}Cl_{2} = 2.23 \times 10^{-3} moles

           1 mole of C_{2}H_{4}Cl_{2} = 22.4 L

   2.23 \times 10^{-3} moles = 22.4 \times 2.23 \times 10^{-3} moles        

                                = 50 ml of product

Thus, we can conclude that 50 ml of products if pressure and temperature are kept constant.

5 0
4 years ago
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8 0
3 years ago
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mr Goodwill [35]

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5 0
3 years ago
A 825 g iron block is heated to 352 degrees C and is placed in an insulated container (of negligible heat capacity) containing 4
Stella [2.4K]

Answer : The final equilibrium temperature of the water and iron is, 537.12 K

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of iron =  560 J/(kg.K)

c_1 = specific heat of water = 4186 J/(kg.K)

m_1 = mass of iron = 825 g

m_2 = mass of water = 40 g

T_f = final temperature of water and iron = ?

T_1 = initial temperature of iron = 352^oC=273+352=625K

T_2 = initial temperature of water = 20^oC=273+20=293K

Now put all the given values in the above formula, we get:

(825\times 10^{-3}kg)\times 560J/(kg.K)\times (T_f-625K)=-(40\times 10^{-3}kg)\times 4186J/(kg.K)\times (T_f-293K)

T_f=537.12K

Therefore, the final equilibrium temperature of the water and iron is, 537.12 K

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