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zysi [14]
3 years ago
14

Match the symbol for each element with the name of the element. 1. hydrogen K 2. helium Mg 3. sodium He 4. magnesium H 5. potass

ium Na
Chemistry
1 answer:
insens350 [35]3 years ago
5 0

Answer:

1. hydrogen - H

2. helium - He

3. sodium - Na

4. magnesium - Mg

5. potassium - K

Explanation:

Hydrogen is the element of group 1 and first period. The atomic number of hydrogen is 1 and the symbol of the element is H.

The electronic configuration of the element hydrogen is:-

1s^1

Helium is the element of group 18 and first period. The atomic number of helium is 2 and the symbol of the element is He.

The electronic configuration of the element helium is:-

1s^2

Sodium is the element of group 1 and third period. The atomic number of sodium is 11 and the symbol of the element is Na.

The electronic configuration of the element sodium is:-

1s^22s^22p^63s^1

Magnesium is the element of group 2 and third period. The atomic number of magnesium is 12 and the symbol of the element is Mg.

The electronic configuration of the element magnesium is:-

1s^22s^22p^63s^2

Potassium is the element of group 1 and forth period. The atomic number of potassium is 19 and the symbol of the element is K.

The electronic configuration of the element potassium is:-

1s^22s^22p^63s^23p^64s^1

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Answer is: sodium (Na) and iodine (I₂).

<span> First ionic bonds in this salt are separeted because of heat: 
</span>NaI(l) → Na⁺(l) + I⁻(l).

Reaction of reduction at cathode(-): Na⁺(l) + e⁻ → Na(l) /×2.

2Na⁺(l) + 2e⁻ → 2Na(l).

Reaction of oxidation at anode(+): 2I⁻(l) → I₂(l) + 2e⁻.

The anode is positive and the cathode is negative.


4 0
3 years ago
At a certain temperature the vapor pressure of pure benzene is measured to be . Suppose a solution is prepared by mixing of benz
Marianna [84]

Answer:

P(C₆H₆) = 0.2961 atm

Explanation:

I found an exercise pretty similar to this, so i'm gonna use the data of this exercise to show you how to do it, and then, replace your data in the procedure so you can have an accurate result:

<em>"At a certain temperature the vapor pressure of pure benzene (C6H6) is measured to be 0.63 atm. Suppose a solution is prepared by mixing 79.2 g of benzene and 115. g of heptane (C7H16) Calculate the partial pressure of benzene vapor above this solution. Round your answer to 2 significant digits. Note for advanced students: you may assume the solution is ideal".</em>

<em />

Now, according to the data, we want partial pressure of benzene, so we need to use Raoul's law which is:

P = Xₐ * P°    (1)

Where:

P: Partial pressure

Xₐ: molar fraction

P°: Vapour pressure

We only have the vapour pressure of benzene in the mixture. We need to determine the molar fraction first. To do this, we need the moles of each compound in the mixture.

To get the moles:   n = m / MM

To get the molar mass of benzene (C₆H₆) and heptane (C₇H₁₆), we need the atomic weights of Carbon and hydrogen, which are 12 g/mol and 1 g/mol:

MM(C₆H₆) = (12*6) + (6*1) = 78 g/mol

MM(C₇H₁₆) = (7*12) + (16*1) = 100 g/mol

Let's determine the moles of each compound:

moles (C₆H₆) = 79.2 / 78 = 1.02 moles

moles (C₇H₁₆) = 115 / 100 = 1.15 moles

moles in solution = 1.02 + 1.15 = 2.17 moles

To get the molar fractions, we use the following expression:

Xₐ = moles(C₆H₆) / moles in solution

Xₐ = 1.02 / 2.17 = 0.47

Finally, the partial pressure is:

P(C₆H₆) = 0.47 * 0.63

<h2>P(C₆H₆) = 0.2961 atm</h2>

Hope this helps

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