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3 years ago

Elements found to the left of the metalloids on the periodic table display which properties? Select all that apply.

2 answers:

8 0

Metals tend to be good conductors, they tend to form cations, They are lustrous, Can't be broken easily, High density, ductile, tend to gain valence electrons and so on.

Katyanochek1 [597]3 years ago

4 0

They are solid

They are shiny, good conductors of electricity and heat.

They are ductile

They are malleable

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6.) The bond between which two atoms is most polar?*

Feliz [49]

Answer:

O-H bond

Explanation:

Let us work out the electronegativity difference between the elements in each bond in order to decide which of them is most polar.

For the C-O bond

2.55 - 2.2 =0.35

For the F-F bond

3.98 - 3.98 = 0

For the O-H bond

3.44 - 2.2 = 1.24

For the N-H bond

3.04 - 2.2 = 0.84

The O-H bond has the highest electronegativity difference, hence it is he most polar bond.

4 0

2 years ago

Given LiCl, NaCl, KCl, MgCl2, and SrCl2, arrange the compounds in order of increasing lattice energy.

Oksanka [162]

Answer:- $KCl$

Explanations:- Lattice energy depends on two factors, charge and size.

High charge and small size gives higher lattice energy where as low charge and bigger size gives lower lattice energy.

in LiCl, NaCl and KCl, the anion is same and also the charges for Li, Na and K are also same. The deciding factor here is the size of cations. Since the size increases as we move down a group, the order of size of these three atoms is Li<Na<K.

The order of lattice energy is exactly opposite as it's increases as the size decreases.

Now, if we look at magnesium chloride and strontium chloride then again the anion is common but the metals have higher charge as compared to the alkali metals(Li, Na and K). So, lattice energy values must be higher for these two compounds. If we compare Mg and Sr then size of Mg is smaller and so the lattice energy would be greater for this.

Hence, the increasing order of lattice energy is $KCl$ .

4 0

2 years ago

Read 2 more answers

What are kind and number of atoms in ring portion in the haworth structure of glucose

monitta

Answer:

The six member ring and the position of the -OH group on the carbon (#4) identifies glucose from the -OH on C # 4 in a down projection in the Haworth structure). Fructose is recognized by having a five member ring and having six carbons, a hexose.

3 0

2 years ago

Read 2 more answers

What mass, in grams, of carbon dioxide gas can be produced by complete combustion of 24.0 grams of carbon?

Elan Coil [88]

Jhuftghhhbbrdcbjkkjhuuuyyggfggg

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2 years ago

Under standard-state conditions, which of the following species is the best reducing agent? a. Ag+ b. Pb c. H2 d. Ag e. Mg2+

eimsori [14]

Answer: The correct answer is Option b.

Explanation:

Reducing agents are defined as the agents which help the other substance to get reduced and itself gets oxidized. They undergo oxidation reaction.

$X\rightarrow X^{n+}+ne^-$

For determination of reducing agents, we will look at the oxidation potentials of the substance. Oxidation potentials can be determined by reversing the standard reduction potentials.

For the given options:

Option a: $Ag^+$

This ion cannot be further oxidized because +1 is the most stable oxidation state of silver.

Option b: $Pb$

This metal can easily get oxidized to $Pb^{2+}$ ion and the standard oxidation potential for this is 0.13 V

$Pb\rightarrow Pb^{2+}+2e^-;E^o_{(Pb/Pb^{2+})}=+0.13V$

Option c: $H_2$

This metal can easily get oxidized to $H^{+}$ ion and the standard oxidation potential for this is 0.0 V

$H_2\rightarrow 2H^++2e^-;E^o_{(H_2/H^{+})}=0.0V$

Option d: $Ag$

This metal can easily get oxidized to $Ag^{+}$ ion and the standard oxidation potential for this is -0.80 V

$Ag\rightarrow Ag^{+}+e^-;E^o_{(Ag/Ag^{+})}=-0.80V$

Option e: $Mg^{2+}$

This ion cannot be further oxidized because +2 is the most stable oxidation state of magnesium.

By looking at the standard oxidation potential of the substances, the substance having highest positive $E^o$ potential will always get oxidized and will undergo oxidation reaction. Thus, considered as strong reducing agent.

From the above values, the correct answer is Option b.

8 0

3 years ago