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MaRussiya [10]
2 years ago
11

An atom of a mystery element contains 6 protons, 6 electrons, and 8 neutrons.

Chemistry
1 answer:
Leno4ka [110]2 years ago
8 0

Answer:

14

Explanation:

Mass number is just the value of the protons and neutrons added, in this case its 14

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Calculate the freezing point of a solution 1.25 g benzene (C6H6) in 125 g of chloroform (CHCl3).
posledela

Answer:

The freezing point for the solution is -64.09°C

Explanation:

This problem can be solved, by the freezing point depression. This colligative problem shows, that the freezing point of a solution is lower than the freezing point of pure solvent.

ΔT = Kf.  m

ΔT = T° freezing pure solvent - T° freezing solution

Kf = Cryscopic constant, for chloroform is 4.68

T°freezing pure solvent = -63.5°C

m is mol/kg of solvent → molality

Let's determine the moles of benzene

1.25 g / 78 g/mol = 0.0160 mol

Let's convert the mass of solvent to kg

125 g . 1kg / 1000 g = 0.125 kg

m = 0.0160 mol / 0.125 kg → 0.128 m

Let's go to the formula to replace the data

-63.5°C - T° freezing solution = 4.68 °C/m . 0.128 m

T° freezing solution = - (4.68 °C/m . 0.128 m + 63.5°C)

T° freezing solution = - 64.09°C

3 0
3 years ago
Why do the elements in a group all behave similarly
Svetradugi [14.3K]

Answer:

See the explanation below, please.

Explanation:

The elements of the periodic table that belong to the same group (each column) have similar physical and chemical properties. This is because they have the same number of electrons in their last electronic layer.

Example of electronic configuration of elements of GROUP IA:

Hydrogen: 1s ^ 1

Lithium: 1s ^ 2 2s ^ 1

6 0
3 years ago
3. A diamond contains 0.090 moles of carbon What is the mass of the diamond?
Nady [450]

Answer:

1.0809

Explanation:

3 0
2 years ago
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Answer:

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5 0
3 years ago
What are the products obtained in the electrolysis of molten nai?
yanalaym [24]
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
2 years ago
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