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3 years ago

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A student has a sample of isopropanol (C3H7OH) that has a mass of 78.6 g. The molar mass of isopropanol is 60.1 g/mol. How many

moles of isopropanol are contained in this sample?

1 answer:

vivado [14]3 years ago

5 0

78.6 g (1mol/60.1 g)= 1.31 moles of isopropanol

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Suppose you begin with an unknown volume of 8.61 m h2so4 and add enough water to make 5.00*102 ml of a 1.75 m h2so4 solution. wh

olga55 [171]

<span>1.02x10^2 ml Since molarity is defined as moles per liter, the product of the molarity and volume will remain constant as mole solvent is added. So let's set up an equality to express this m0*v0 = m1*v1 where m0, v0 = molarity and volume of original solution m1, m1 = molarity and volume of final solution. Solve for v0, then substitute the known values and calculate: m0*v0 = m1*v1 v0 = (1.75 M * 500 ml)/8.61 M v0 = (1.75 M * 500 ml)/8.61 M V0 = 101.6260163 Rounding to 3 significant figures gives 102 ml. So the original volume of the 8.61 M H2SO4 solution was 102 ml or 1.02x10^2 ml.</span>

3 0

3 years ago

Which of the following changes would have no effect on the equilibrium position of the reaction below? 2 NOBR (g) 2 NO (g)+ Br2

nasty-shy [4]

Answer:

D) cutting the concentrations of both NOBr and NO in half

Explanation:

The equilibrium reaction given in the question is as follows -

2NOBr ( g ) ↔ 2NO ( g ) + Br₂ ( g )

The equilibrium constant for the above reaction can be written as -

K = [ NO ]² [ Br₂ ] / [ NOBr ] ²

Therefore from the condition given in the question , the changes that will not affect the equilibrium will be , reducing the concentration of both NOBr and NO to half ,

Hence ,

the new concentrations are as follows -

[ NoBr ] ' = 1/2 [ NoBr ]

[ NO ] ' = 1/2 [ NO ]

Hence the new equilibrium constant equation can be written as -

K ' = [ NO ] ' ² [ Br₂ ] / [ NOBr ] ' ²

Substituting the new concentration terms ,

K ' = 1/2 [ NO ] ² [ Br₂ ] / 1/2 [ NoBr ] ²

K ' = 1/4 [ NO ] ² [ Br₂ ] / 1/4 [ NoBr ] ²

The value of 1 / 4 in the numerator and the denominator is cancelled -

K ' = [ NO ] ² [ Br₂ ] / [ NoBr ] ²

Hence ,

K' = K

5 0

4 years ago

The quantum numbers for the last electron placed in three elements are listed below. Which of these is(are) NOT correct? Er (4 3

UNO [17]

Answer:

The three elements Erbium, Thallium and Osmium have incorrect quantum numbers for the last electron placed.

Explanation:

The 4 quantum numbers are (<em>n,l,ml,ms</em>):

<em>n</em> (Principal quantum number): it is the <u>number of the shell (level)</u> where the electron is placed.
<em>l </em>(Angular momentum quantum number or Secondary): it represents the <u>sublevel where the electron is</u> placed. There are 4 subleves: s, p d and f so secondary quantum number can take the number 0 (s), 1 (p), 2 (d) or 3 (f) depending on which sublevel the electron is placed.
<em>ml</em> (Magnetic quantum number): it represents the <u>spatial orientation</u> of the electron <u>in respect of the sublevel the electron</u> is placed. For example: if the electron occupies the <em>s sublevel</em> the magnetic number will be <em>0</em>, if the electron occupies the <em>p sublevel</em> the magnetic number could be <em>-1,0,1</em>, if the electron occupies the <em>d sublevel</em> the magnetic number could be <em>-2,-1,0,1,2</em> and if the electron occupies the <em>f sublevel</em> the magnetic number could be <em>-3,-2,-1,0,1,2,3</em>. You can see this in the attachment related to the correct sublevel for the example.

<em>ms</em> (Spin quantum number): this number represents the possible rotation of the electron so it could be 1/2 (which is represented by an up arrow) or -1/2 (represented by an down arrow).

Let's analyze the last electron of each element. You can see the attachment for better understanding. The last electron it is represented with orange color.

- Erbium:

This element has 68 electrons so following the Moeller's Diagram to fill the the electronic configuration, we found that the last electron of Erbium it is in the <u>4th level </u>(shell), in the <u>f sublevel</u>. As Erbium has 12 electrons in the f sublevel, it is necessary to follow the Hund's rule (electrons must be placed singly in every sublevel before place a parallel electron) to placed correctly all of them. Finally, the last electron of Erbium stays in the middle of the sublevel and it is represented by a down arrow so the correct quantum numbers in the Erbium element are (4,3,1,-1/2).

- Thallium:

This element has 81 electrons and following the Moeller's Diagram, we found that it last electron it is in the <u>6th level</u>, in the <u>p sublevel</u>. As Thallium has 1 electron in the p sublevel, it is placed singly in the sublevel. So the last electron of Thallium it is represented by an up arrow so the correct quantum numbers in the Thallium element are (6,1,-1,1/2).

- Osmium:

Osmium has 76 electrons and following the steps that we did with we the other elements, we noticed that its last electron it is in the <u>5th level</u>, in the <u>d sublevel</u>. Following the Hund's rule the last electron of Osmium has a magnetic quantum number of -2 and its spin quantum number is -1/2, so the quantum numbers in the Osmium element are (5,2,-2,-1/2).

<u>Note:</u>

- Remember that the <em>s sublevel</em> has place for 2 electrons, the <u>p sublevel</u> has place for 6 electrons, the <u>d sublevel</u> has place for 10 electrons and the<em> f sublevel</em> has place for 14 electrons.

3 0

3 years ago

11. Power can be defined as (2 points)

Ierofanga [76]

Answer:

the energy required to do work

4 0

3 years ago

What is the net ionic equation for the reaction that occurs when aqueous solutions of koh and zncl2 are mixed?

Mademuasel [1]

Answer:

Net ionic equation:

Zn²⁺(aq) + 2OH⁻(aq) → Zn(OH)₂(s)

Explanation:

Chemical equation:

ZnCl₂ + KOH → KCl + Zn(OH)₂

Balanced chemical equation:

ZnCl₂ + 2KOH → 2KCl +Zn(OH)₂

Ionic equation;

Zn²⁺(aq) + 2Cl⁻(aq) + 2K⁺(aq) + 2OH⁻(aq) → 2K⁺(aq) + 2Cl⁻(aq) +Zn(OH)₂(s)

Net ionic equation:

Zn²⁺(aq) + 2OH⁻(aq) → Zn(OH)₂(s)

The K⁺ and Cl⁻ are spectator ions that's why these are not written in net ionic equation. The Zn(OH)₂ can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

5 0

3 years ago