CH3CH2CH(Cl)CHO iupac name?

Salt in crude oil must be removed before the oil undergoes processing in a refinery. The

irina1246 [14]

Answer:

$\large \boxed{0.64 \, \%}$

Explanation:

Assume you are using 1 L of water.

Then you are washing 4 L of salty oil.

1. Calculate the mass of the salty oil

Assume the oil has a density of 0.86 g/mL.

$\text{Mass of oil} = \text{4000 mL} \times \dfrac{\text{0.86 g}}{\text{1 mL}} = \text{3440 g}$

2. Calculate the mass of salt in the salty oil

$\text{Mass of salt} = \text{3440 g} \times \dfrac{\text{5 g salt}}{\text{100 g oil}} = \text{172 g salt}$

3. Calculate the mass of salt in the spent water

$\text{Mass of salt} = \text{1000 g water} \times \dfrac{\text{15 g salt}}{\text{100 g water}} = \text{150 g salt}$

4. Mass of salt remaining in washed oil

Mass = 172 g - 150 g = 22 g

5. Concentration of salt in washed oil

$\text{Concentration} = \dfrac{\text{22 g}}{\text{3440 g}} \times 100 \, \% = \mathbf{0.64 \, \%}\\\\\text{The concentration of salt in the washed oil is $\large \boxed{\mathbf{0.64 \, \%}}$}$

3 0

2 years ago

The highest principal enegy level of period 2 elements is 2. Period 3 elements all have six 3p electrons. Period 4 elements have

Dmitrij [34]

Explanation:

The highest principal energy level of period 2 elements is 2. (True)

The highest principle energy level of elements of any period is equal to period number. So the given statement is true.

Period 3 elements all have six 3p electrons – (False)

For example, sodium and sulfur are period 3 elements but do not contain six 3p electrons. Only Argon of period possesses six p-electrons.

Period 4 elements have an inner electron configuration of [Ar]- (True)

Atomic number of elements belonging to period 4 have atomic number more than 18 (atomic number of Ar). So they must have an inner electron configuration of [Ar]

The valence electrons of group 5A elements are in the 6s subshell – [FALSE]

Elements of group 5A are nitrogen, arsenic, antimony and bismuth. They are p-block elements. Therefore, valence electrons are present in p-subshell.

Group 8A elements have full outer principal s and p subshells –(True)

Group 8A elements are also known as noble gas element. Octets of noble gases are complete and hence, have full outer principal s and p subshells.

The valence electrons of group 2A elements are in an s subshell – (True)

Elements of group 2A are barium, magnesium, calcium, etc. They belong to s-block elements. Therefore, their valence electrons will be in s-subshell.

The highest principal energy level of period 3 elements is 4 – (False)

The highest principle energy level of elements of any period is equal to period number.

For example, sodium is present in 3rd period, so its principle quantum number will be 3.

Period 5 elements have an inner electron configuration of [Xe] – (False)

Period 5 elements have an inner electron configuration of [Kr]. The first member of this period is rubidium. 37 and that of Kr is 36.

6 0

2 years ago

Which of the following contains maximum number of atoms? a) 2.0g hydrogen b) 2.0g oxygen c) 2.0g nitrogen d) 2.0g methane

Fynjy0 [20]

A) 2.0g hydrogen is your answer

7 0

2 years ago

10.000g of boron (B) combines with hydrogento form 11.554g of a pure compound. What is the empirical formula of this compound?

Citrus2011 [14]

Answer:

B3H5

Explanation:

The law of conservation of mass states that matter in an closed system is neither created nor destroyed by physical transformations or chemical reactions but changes from one form to the other.

That is, the sum of masses of the reactants = The sum of masses of the product

10.00g of Boron + x grams of Hydrogen = 11.55g of the product

Mass of hydrogen = 11.55 - 10.00 = 1.55g

Molar mass of Boron = 10.811g

Molar mass of Hydrogen = 1.00784g

Number of moles of Boron = (mass of Boron)/(molar mass of Boron) = 10/10.811 = 0.9249 mols

Number of moles of Hydrogen = (mass of Hydrogen)/(molar mass of Hydrogen) = 1.55/1.00784 =1.5379mols

0.9249 mols of Boron combines with 1.5379mols of Hydrogen

Dividing both sides mols by 0.9249 gives

1 mole of Boron combines with 1.66266 mols of Hydrogen

Converting 1.66266 to fractions we have 1.66266 approximately 5/3

or 1 mole of Boron combines with 5/3 moles of Hydrogen

Multiplying both sides by 3 we have

3 moles of Boron combines with 5 moles of Hydrogen

Molecular formula of the compound is

B3H5

4 0

2 years ago

A potential energy diagram is shown. A graph is plotted with Potential Energy in KJ on the y axis and Reaction Pathway on the x

ahrayia [7]

Answer:

35 KJ.

Explanation:

The activation energy is the minimum energy that must be overcome for a reaction to take place.

In the diagram given above, the activation energy lies between the energy of the reactants and that at the peak.

Thus we can calculate the activation energy as follow:

Energy of reactants = 30 KJ

Energy at the peak = 65 KJ

Activation energy =..?

Activation energy = Energy at the peak – Energy of reactants

Activation energy = 65 – 30

Activation energy = 35 KJ

Therefore, the activation energy of th reaction is 35 KJ

3 0

2 years ago