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3 years ago

How many milliliters of a 3.0 M HCL solution are required to make 250.0 millimeters of 1.2 M HCL?

1 answer:

7 0

The problem above can be solved using M1V1=M2V2 where M1 is the concentration of the concentrated, V1 is the volume of the concentrated solution, M2 is the concentration of the Dilute Solution, V2 is the Volume of the dilute solution. Hence,

(3.0 M)(V2)=(250 mL)(1.2M)
V2 (3.0)= 300
V2= 100 mL

Therefore, you need 100 mL of 3.0 M HCl to form a 250 mL of 1.2 M HCl.

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F a shell can hold a maximum of 32 electrons, what is the value of n? 2 3 4 5

Arte-miy333 [17]

Answer: 4

Explanation:

Principle Quantum Numbers: This quantum number describes the size of the orbital. It is represented by n.

Azimuthal Quantum Number: This quantum number describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...

Magnetic Quantum Number: This quantum number describes the orientation of the orbitals. It is represented as $m_l$ . The value of this quantum number ranges from $(-l\text{ to }+l)$ . When l = 2, the value of $m_l$ will be -2, -1, 0, +1, +2.

Given : a f subshell, thus l = 3 , Thus the subshells present would be 3, 2, 1, 0 and thus n will have a value of 4.

Also electrons give are 32.

The formula for number of electrons is $2n^2$ .

$2n^2=32$

$n=4$

Thus principal quantum no will be n= 4.

8 0

2 years ago

Read 2 more answers

In Part III, the phenolphthalein indicator is used to monitor the equilibrium shifts of the ammonia/ammonium ion system. The phe

AnnZ [28]

The pink color in the solution fades. Some of the colored indicator ion converts to the colorless indicator molecule.

<h3>Explanation</h3>

What's the initial color of the solution?

$\text{NH}_4\text{Cl}$ is a salt soluble in water. $\text{NH}_4\text{Cl}$ dissociates into ions completely when dissolved.

$\text{NH}_4\text{Cl} \; (aq)\to {\text{NH}_4}^{+} \; (aq) +{\text{Cl}}^{-} \; (aq)$ .

The first test tube used to contain $\text{NH}_4\text{OH}$ . $\text{NH}_4\text{OH}$ is a weak base that dissociates partially in water.

$\text{NH}_4\text{OH} \; (aq) \rightleftharpoons {\text{NH}_4}^{+} \;(aq)+ {\text{OH}}^{-} \; (aq)$ .

There's also an equilibrium between $\text{OH}^{-}$ and ${\text{H}_3\text{O}}^{+}$ ions.

${\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l)$ .

$\text{OH}^{-}$ ions from $\text{NH}_4\text{OH}$ will shift the equilibrium between $\text{OH}^{-}$ and ${\text{H}_3\text{O}}^{+}$ to the right and reduce the amount of ${\text{H}_3\text{O}}^{+}$ in the solution.

The indicator equilibrium will shift to the right to produce more ${\text{H}_3\text{O}}^{+}$ ions along with the colored indicator ions. The solution will show a pink color.

What's the color of the solution after adding NH₄Cl?

Adding $\text{NH}_4\text{Cl}$ will add to the concentration of ${\text{NH}_4}^{+}$ ions in the solution. Some of the ${\text{NH}_4}^{+}$ ions will combine with $\text{OH}^{-}$ ions to produce $\text{NH}_4\text{OH}$ .

The equilibrium between $\text{OH}^{-}$ and ${\text{H}_3\text{O}}^{+}$ ions will shift to the left to produce more of both ions.

${\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l)$

The indicator equilibrium will shift to the left as the concentration of ${\text{H}_3\text{O}}^{+}$ increases. There will be less colored ions and more colorless molecules in the test tube. The pink color will fade.

7 0

3 years ago

In the following equation:

Luda [366]

Answer:

FeCl₃

Explanation:

4FeCl₃ + 3O₂ => 2Fe₂O₃+ 6Cl₂

Given => 7moles 9moles

A simple way to determine which reagent is the limiting reactant is to convert all given data to moles then divide by the respective coefficients of the balanced equation. The smaller value will be the limiting reactant.

4FeCl₃ + 3O₂ => 2Fe₂O₃+ 6Cl₂

Given => 7/4 = 1.75* 9/3 = 3

*Smaller value => FeCl₃ is limiting reactant.

NOTE: However, when working problems, one must use original mole values given.

7 0

3 years ago

View Available Hint(s) Check all that apply. Hydrogen bonding occurs when a hydrogen atom is covalently bonded to an NN, OO, or

Sauron [17]

Answer:

Hydrogen bonding occurs when a hydrogen atom is covalently bonded to an NN, OO, or FF atom.

A hydrogen atom acquires a partial positive charge when it is covalently bonded to an FF atom.

A hydrogen bond is possible with only certain hydrogen-containing compounds.

Explanation:

A hydrogen bond does not occur in all hydrogen containing compounds. Hydrogen bonds only occur in those compounds where hydrogen is bonded to a highly electronegative element such as fluorine, oxygen or nitrogen.

In a hydrogen bonded specie, hydrogen acquires a partial positive charge and the electronegative element acquires a partial negative charge which extends throughout the molecule.

7 0

3 years ago

Repulsion of electrons within two interacting molecules produces changes in electron distribution. This change in electron distr

ycow [4]

Answer:

Explanation:

This explains how two noble gases molecules can have an attractive force between them.

This force is called as van dar Waals forces.

It plays a fundamental role in fields in as diverse as supramolecular chemistry structural biology .

If no other forces are present, the point at which the force becomes repulsive rather than attractive as two atoms near one another is called the van der Waals contact distance. This results from the electron clouds of two atoms unfavorably coming into contact.[1] It can be shown that van der Waals forces are of the same origin as the Casimir effect, arising from quantum interactions with the zero-point field.[2] The resulting van der Waals forces can be attractive or repulsive.[3] It is also sometimes used loosely as a synonym for the totality of intermolecular forces.[4] The term includes the force between permanent dipoles (Keesom force), the force between a permanent dipole and a corresponding induced dipole (Debye force), and the force between instantaneously induced dipoles

6 0

2 years ago