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3 years ago

How many atoms of chlorine are in the carbon tetrachloride molecule?

Chemistry

2 answers:

zaharov [31]3 years ago

7 0

Answer:

4 atoms of Chlorine

Explanation:

This is actually easy to explain,

First we have here a type of nomenclature. This one is the sistematic nomenclature, and begins by naming the number of atoms that one element has, beggining for the non metal first, and then the metal or the non metal acting like a metal. In this case, the Carbon is acting like a metal.

The number of atoms are named by a prefix of the number. Each number has a determined prefix. Here are some of them:

one = 1 = mono

two = 2 = bi or di

three = 3 = tri

Four = 4 = tetra

Five = 5 = penta

Six = 6 = Hexa

So tetrachloride, means that we have 4 atoms of chlorine in the molecule and the molecule is this one: CCl4

kap26 [50]3 years ago

5 0

The answer is 4. The user below is correct.

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Diano4ka-milaya [45]

Answer:

$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

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Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

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$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$

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3 years ago

A piece of mental with a mass of 106 g is placed in a 50 mL graduated cylinder.The volume rises from 20 mL to 31 mL. What is the

garik1379 [7]

Answer:

9.64g/mL

Explanation:

Given parameters:

Mass of the metal = 106g

Volume of cylinder = 50mL

Volume difference = 31mL - 20mL = 11mL

Unknown:

Density of the metal = ?

Solution:

To find the density of the metal, we use;

Density = $\frac{mass}{volume}$

Density = $\frac{106}{11}$ = 9.64g/mL

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A gas within a piston–cylinder assembly undergoes an isothermal process at 400 K during which the change in entropy is −0.3 kJ/K

Karolina [17]

Answer:

W = -120 KJ

Explanation:

Since the piston–cylinder assembly undergoes an isothermal process, then the temperature is constant.

Thus; T1 = T2 = 400K

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Formula for change in entropy is written as;

ΔS = Q/T

Where Q is amount of heat transferred.

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Q = ΔU + W

Where;

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Now, since it's an ideal gas model, the change in internal energy is expressed as;

ΔU = m•C_v•ΔT

Where;

m is mass

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ΔT is change in temperature

Now, since it's an isothermal process where temperature is constant, then;

ΔT = T2 - T1 = 0

Thus;

ΔU = m•C_v•ΔT = 0

ΔU = 0

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Answer:

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