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4 years ago

Which metal atom below cannot form a cation of several different charges?

1 answer:

4 0

Missing question: Cr, Mn, Fe, Ba, Cu.
Answer is: Ba (barium).
Barium is metal from second group of periodic table of elements. Barium looses two electrons and has oxidation number +2.
Chromium has different oxidation numbers, usually +3 and +6, iron has oxidation numbers +2 and +3, copper +1 and +2 in their compounds.

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SOLVE A 4.80-gram piece of magnesium displaces 2.76 mL of water when it is placed in a graduated cylinder. What is the density o

Stels [109]

Answer:

Density = 1.74 gm/mL

Explanation:

Given that:

Mass of piece of Magnesium = 4.80 grams

The volume displaced by this piece = 2.76mL

Let us learn about a property:

Volume of water displaced by a substance when immersed in the water is equal to the Volume of the substance itself.

So, volume of the piece of Magnesium = 2.76 mL

To find:

Density of magnesium = ?

Solution:

Let us have a look at the formula for Density of a Substance:

Density of a substance is given as the ratio of mass of substance and the volume of that mass of substance.

i.e.

$Density = \dfrac{Mass}{Volume}$

Putting the given values:

$Density = \dfrac{4.80}{2.76}\\\Rightarrow \bold{Density = 1.74\ gm/mL}$

So, the answer is:

Density of Magnesium = 1.74 gm/mL

4 0

3 years ago

Be sure to answer all parts. The sulfate ion can be represented with four S―O bonds or with two S―O and two S═O bonds. (a) Which

VMariaS [17]

Answer:

a) Representation - (in attachment)

b) Tetrahedral geometry and

c) $sp^{3}$ hybrid orbitals invovle sigma bonding.

$\pi$ orbitals are formed by the overlapping of d-orbitals of sulfur with p-orbitals of oxygen.

Explanation:

Representation in attachment.

The overall charge in the both structures are -2. The structure (b) is favored by the resonance structures since the formal charge with in the species are remained.

Therefore, structure (b) is the better representation of sulfate ion.

In sulfate ion, sulfur atom is attached with four different oxygen atoms. According to the VSEPR theory , the sufate ion has tetrahedral geometry.

There is four sigma bonds and zero lone pairs present on the central metal atom.

Hence, the hybridization of the sulfur atom is $sp^{3}$

The s and p orbitals of sulfur invovles hybridization and form sigma bonds.The orbitals of sulfur involves $\pi$ bonding.

Therefore, $\pi$ bonds are formed by the overlapping of d-orbitals of sulfur with p-orbitals of oxygen.

5 0

4 years ago

When H+ forms a Bond with H2O to form the Hydronium ion H3 plus this bond is called a coordinate covalent bond because

NISA [10]

Answer:

Because both the bonding electrons come from the oxygen atom. Explanation: A coordinate covalent bond is formed when both the bonding electrons are coming from the same atom

Explanation:

7 0

3 years ago

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.210 M HClO(aq) with 0.210 M KOH(aq).

Degger [83]

a) before addition of any KOH :

when we use the Ka equation & Ka = 4 x 10^-8 :

Ka = [H+]^2 / [ HCIO]

by substitution:

4 x 10^-8 = [H+]^2 / 0.21

[H+]^2 = (4 x 10^-8) * 0.21

= 8.4 x 10^-9

[H+] = √(8.4 x 10^-9)

= 9.2 x 10^-5 M

when PH = -㏒[H+]

PH = -㏒(9.2 x 10^-5)

= 4

b)After addition of 25 mL of KOH: this produces a buffer solution

So, we will use Henderson-Hasselbalch equation to get PH:

PH = Pka +㏒[Salt]/[acid]

first, we have to get moles of HCIO= molarity * volume

=0.21M * 0.05L

= 0.0105 moles

then, moles of KOH = molarity * volume

= 0.21 * 0.025

=0.00525 moles

∴moles HCIO remaining = 0.0105 - 0.00525 = 0.00525

and when the total volume is = 0.05 L + 0.025 L = 0.075 L

So the molarity of HCIO = moles HCIO remaining / total volume

= 0.00525 / 0.075

=0.07 M

and molarity of KCIO = moles KCIO / total volume

= 0.00525 / 0.075

= 0.07 M

and when Ka = 4 x 10^-8

∴Pka =-㏒Ka

= -㏒(4 x 10^-8)

= 7.4

by substitution in H-H equation:

PH = 7.4 + ㏒(0.07/0.07)

∴PH = 7.4

c) after addition of 35 mL of KOH:

we will use the H-H equation again as we have a buffer solution:

PH = Pka + ㏒[salt/acid]

first, we have to get moles HCIO = molarity * volume

= 0.21 M * 0.05L

= 0.0105 moles

then moles KOH = molarity * volume
= 0.22 M* 0.035 L

=0.0077 moles

∴ moles of HCIO remaining = 0.0105 - 0.0077= 8 x 10^-5

when the total volume = 0.05L + 0.035L = 0.085 L

∴ the molarity of HCIO = moles HCIO remaining / total volume

= 8 x 10^-5 / 0.085

= 9.4 x 10^-4 M

and the molarity of KCIO = moles KCIO / total volume

= 0.0077M / 0.085L

= 0.09 M

by substitution:

PH = 7.4 + ㏒( 0.09 /9.4 x 10^-4)

∴PH = 8.38

D)After addition of 50 mL:

from the above solutions, we can see that 0.0105 mol HCIO reacting with 0.0105 mol KOH to produce 0.0105 mol KCIO which dissolve in 0.1 L (0.5L+0.5L) of the solution.

the molarity of KCIO = moles KCIO / total volume

= 0.0105mol / 0.1 L

= 0.105 M

when Ka = KW / Kb

∴Kb = 1 x 10^-14 / 4 x 10^-8

= 2.5 x 10^-7

by using Kb expression:

Kb = [CIO-] [OH-] / [KCIO]

when [CIO-] =[OH-] so we can substitute by [OH-] instead of [CIO-]

Kb = [OH-]^2 / [KCIO]

2.5 x 10^-7 = [OH-]^2 /0.105

∴[OH-] = 0.00016 M

POH = -㏒[OH-]

∴POH = -㏒0.00016

= 3.8
∴PH = 14- POH

=14 - 3.8

PH = 10.2

e) after addition 60 mL of KOH:

when KOH neutralized all the HCIO so, to get the molarity of KOH solution

M1*V1= M2*V2

when M1 is the molarity of KOH solution

V1 is the total volume = 0.05 + 0.06 = 0.11 L

M2 = 0.21 M

V2 is the excess volume added of KOH = 0.01L

so by substitution:

M1 * 0.11L = 0.21*0.01L

∴M1 =0.02 M

∴[KOH] = [OH-] = 0.02 M

∴POH = -㏒[OH-]

= -㏒0.02

= 1.7

∴PH = 14- POH

= 14- 1.7

= 12.3

8 0

3 years ago

Read 2 more answers

What characteristics of fish might allow them to survive

uranmaximum [27]

The fact that they lay like 300 eggs. And maybe that they are fast.

3 0

4 years ago

Read 2 more answers