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3 years ago

How many ml of a 4.0 M Mg(OH)2 are needed to make 750 ml of 0.2 M Mg(OH)2???

Chemistry

1 answer:

cupoosta [38]3 years ago

8 0

Answer:

37.5 mL of a 4.0 M Mg(OH)₂ are needed to make 750 mL of 0.2 M Mg(OH)₂

Explanation:

Dilution is the reduction in concentration of a chemical in a solution. Dilution consists of lowering the amount of solute per unit volume of solution.

In other words, dilution is the procedure followed to prepare a less concentrated solution from a more concentrated one, and consists of adding solvent to an existing solution. The quantity or mass of the solute is not changed but only that of the solvent.

In dilutions the expression is used:

Ci*Vi = Cf*Vf

where:

Ci: initial concentration
Vi: initial volume
Cf: final concentration
Vf: final volume

In this case:

Ci: 4 M
Vi: ?
Cf: 0.2 M
Vf: 750 mL

Replacing:

4 M* Vi= 0.2 M* 750 mL

Solving:

$Vi=\frac{0.2 M*750 mL}{4 M}$

Vi= 37.5 mL

37.5 mL of a 4.0 M Mg(OH)₂ are needed to make 750 mL of 0.2 M Mg(OH)₂

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slader Consider the following reactions: A. uranium-238 emits an alpha particle; B. plutonium-239 emits an alpha parti- cle; C.

pishuonlain [190]

Answer:

For A: The equation is $_{92}^{238}\textrm{U}\rightarrow _{90}^{234}\textrm{Th}+_2^4\alpha$

For B: The equation is $_{94}^{239}\textrm{Pu}\rightarrow _{92}^{235}\textrm{U}+_2^4\alpha$

For C: The equation is $_{90}^{239}\textrm{Th}\rightarrow _{91}^{235}\textrm{Pa}+_{-1}^0\beta$

Explanation:

Alpha decay process is the process in which nucleus of an atom disintegrates into two particles. The first one which is the alpha particle consists of two protons and two neutrons. This is also known as helium nucleus. The second particle is the daughter nuclei which is the original nucleus minus the alpha particle released.

$_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha$

Beta decay process is defined as the process the neutrons get converted into an electron and a proton. The released electron is known as the beta particle. In this process, the atomic number of the daughter nuclei gets increased by a factor of 1 but the mass number remains the same.

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta$

For A: Uranium-238 emits an alpha particle

The nuclear equation for this process follows:

$_{92}^{238}\textrm{U}\rightarrow _{90}^{234}\textrm{Th}+_2^4\alpha$

For B: Plutonium-239 emits an alpha particle

The nuclear equation for this process follows:

$_{94}^{239}\textrm{Pu}\rightarrow _{92}^{235}\textrm{U}+_2^4\alpha$

For C: Thorium-239 emits a beta particle

The nuclear equation for this process follows:

$_{90}^{239}\textrm{Th}\rightarrow _{91}^{235}\textrm{Pa}+_{-1}^0\beta$

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3 years ago

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what i

Ostrovityanka [42]

Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :

[butane]=1.0 M , [isobutane]=2.5 M

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?

Answer:

The equilibrium concentration of each gas:

[Butane] = 1.14 M

[isobutane] = 2.86 M

Explanation:

Butane ⇄ Isobutane

At equilibrium

1.0 M 2.5 M

After addition of 0.50 M of butane:

(1.0 + 0.50) M -

After equilibrium reestablishes:

(1.50-x)M (2.5+x)

The equilibrium expression will wriiten as:

$K_c=\frac{[Isobutane]}{[Butane]}$

$2.5=\frac{2.5+x}{(1.50-x)}$

x = 0.36 M

The equilibrium concentration of each gas:

[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M

[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M

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3 years ago

Explain why the first ionization energy K is less than that of Ca

Ainat [17]

Answer:

The first ionization energy for K is less than Ca because Ca has a larger effective nuclear charge.

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3 years ago

In which of these diatomic molecules would you NOT find an octet of electrons

Semmy [17]

Answer:

The hydrogen molecule is the only one in which can not find an octet of electrons around each atom.

Explanation:

Let's evaluate each case.

1. Nitrogen (N₂):

With Z = 7, nitrogen has the following electronic configuration

1s²

2s² 2p³ → valence electrons

Since its valence electrons are 5, in the molecule one nitrogen atom shares 3 electrons with the other one, and each remains with an electron pair, so each atom has an octet of electrons.

2. Hydrogen (H₂):

With Z = 1, its electronic configuration is:

1s¹ → valence electron

In the molecule, the hydrogen atoms share the only electron they have, so they will have only 2 electrons around. In this diatomic molecule, we can not find an octet.

3. Oxygen (O₂):

Z = 8. Electronic configuration:

1s²

2s² 2p⁴ → valence electrons

In the diatomic molecule, each oxygen atom shares 2 electrons with the other one and remains with 2 pairs of electrons, therefore, each oxygen atom has an octet.

4. Fluorine (F₂)

Z = 9. Electronic configuration:

1s²

2s² 2p⁵ → valence electrons

In this molecule, each fluorine atom shares 1 electron with the other and remains with 3 pairs of electrons, hence, each fluorine atom has an octet of electrons around.

Finally, we can say that the hydrogen molecule is the only one in which can not find an octet of electrons around each atom.

I hope it helps you!

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Answer:

The Only Viable Solution: A New Moon Shot

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4 years ago