Which element is malleable and a good conductor of electricity at STP?(1) argon (3) iodine
(2) carbon (4) silver

Answers

Answer 1

Answer:

Answer: The correct answer is option(4).

Explanation:

Malleability : It is a property of a material that under the pressure they can be drawn into thin sheets.

Conducting: It is a property of a material to allow electric current to easily pass through it.

Since both the above properties are of metals. And from the enlisted elements only one element is metal that is 'silver'.

So, the correct answer is option(4).

Answer 2

Answer:

$\boxed{\left( 4 \right){\text{ silver}}}$ is the element that is malleable and a good conductor of electricity at STP.

Further Explanation:

The representation of all the elements in their increasing atomic numbers is known as the periodic table. Such a representation is done in the form of a table having horizontal rows called periods and vertical columns called groups.

Malleability is the property of substance due to which it can be deformed by applying pressure. The malleable materials can be drawn into thin sheets by the process of rolling or hammering. Metals are malleable in nature because these have the tendency to get deformed under pressure.

Any substance that allows the flow of electric charges through it is said to be a conductor. Conductivity is defined as the property of any conductor to permit the flow of charges through it. Metals are good conductors of heat and electricity due to the presence of free electrons in them. Such electrons act as charge carriers.

Both malleability and conductivity are the properties of metals. Since silver exists as a metal at STP, it is malleable and a good conductor of electricity. But carbon, argon and iodine are non-metals at STP so these cannot exhibit any of the mentioned properties. Therefore option (4) is correct.

Learn more:

Which property do metalloids share with nonmetals? brainly.com/question/802690
Which properties are characteristic of group 2 elements? brainly.com/question/8226316

Answer details:

Grade: High School

Chapter: Periodic classification of elements

Subject: Chemistry

Keywords: element, silver, carbon, iodine, argon, conductor, periodic table, metal, non-metals, malleability, electricity, electrons, charge carriers.

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Explanation:

took the test

Answer:

ITS A JUST FINISHED

Explanation:

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Answers

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A hydrocarbon of unknown structure has the formula C8H8. On catalytic hydrogenation over the Lindlar catalyst, 2 equivalents of hydrogen are absorbed. On hydrogenation over a palladium catalyst, 5 equivalents of H2 are absorbed. a) The degrees of unsaturation in the unknown is(are) ?
b) The number of triple bonds is ?
c) The number of double bonds is ?
d) The number of rings is ?

Answers

a) Degree of unsaturation: 5.

C8H8 corresponds to the saturated alkane C8H18, octane, which has eight hydrogen atoms per molecule. Theoretically, it will take five hydrogen molecules H2 to convert C8H8 to C8H18. C8H8 therefore has and DoU of 5.

b) Number of triple bonds: 2.

Only triple bonds are hydrogenated over a Lindlar catalyst. A triple bond can be converted only to a double bond, but not a single bond, under such settings. Each triple bond would absorb only one molecule of hydrogen.

Two hydrogen molecules are absorbed for each C8H8 molecule. Each molecule therefore contains two triple bonds.

c) Number of double bonds: 1.

Palladium catalysts convert both triple and double bonds to single bonds while preserving any rings. Each triple bond would consume two hydrogen molecules, whereas each double bond would consume one.

The two triple bonds as determined in b) account for the consumption of four out the five hydrogen molecules. One double bond shall be responsible for the other.

d) There is no ring in C8H18.

The molecular formula of C8H18 indicates an DoU of five. Each double bond contributes to the DoU by one, each triple bond by two. The double bond and the two triple bonds would have accounted for all five degrees of unsaturation. There are thus no spare DoU available for the presence of rings.

Final answer:

The described hydrocarbon with formula C8H8 has 4 degrees of unsaturation initially, indicating the presence of 2 triple bonds converted by Lindlar catalyst. The remaining 2 degrees of unsaturation are attributed to a ring structure.

Explanation:

The concept being evaluated here is degrees of unsaturation (also known as the index of hydrogen deficiency) in organic molecules and how they refer to the number of double bonds, triple bonds, or rings that a molecule contains.

At first, we can calculate the Index of Hydrogen Deficiency (IHD) which is calculated using the formula (2C+2+X-N)/2, where C is the number of Carbon atoms, X is the halogens and N is the nitrogens. For a molecule with formula C8H8, this equates to (2×8+2-8)/2 = 4. Therefore, your hydrocarbon has 4 degrees of unsaturation.

Now, considering the information that under Lindlar catalyst, the compound absorbed 2 equivalents of H2, which suggests the presence of 2 triple bonds converted to 2 double bonds. This leaves us with 2 degrees of unsaturation which must be in the form of a ring. However, on hydrogenation over a palladium catalyst, 5 equivalents of H2 indicate that these 2 double bonds from the Lindlar reaction and the ring can convert completely to single bonds. So, In summary: a) Degrees of unsaturation: 4 b) Triple bonds: 2 c) Double bonds: 0 (after Lindlar hydrogenation) d) Number of Rings: 1

Learn more about Degrees of Unsaturation here:

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What type of bond is this combination most likely to form?

Answers

Sodium is a group 1 element with atomic number 1. It has 11 electrons. It is soft reactive metal. It has 1 valence electron.

Fluorine is a group 7 element, a hologen with 7 valence electron. It is a most reactive non metal.

When sodium react with fluorine, ionic bond is formed in the resulting compound sodium fluoride.

One sodium and fluorine each totaling 2 atoms are enough to make the bond.

As the bond is formed, both atoms have octet structure. That is they each have 8 electrons on their outermost shells.

The positive charge on sodium indicates that sodium had lost 1 electron to fluorine atom.

The negative charge on fluorine ion indicates that fluorine atom had gained 1 electron from sodium atom to form negative ion.

The name of the compound is sodium fluoride with formula NaF.

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Which of these 1 M solutions will have the highest pH?(1) NaOH (3) HCl(2) CH3OH (4) NaCl

Answers

Answer:

(1) NaOH

Explanation:

NaOH, sodium hydroxide is a base. HCl, hydrochloric acid is an acid. CH3OH, methanol is an alcohol. NaCl, sodium chloride is salt.

pH below 7.0 is acidic, pH above 7.0 is basic, and pH at 7.0 is neutral.

Since NaOH is the only base in among the given compounds, its pH will above 7.0. Therefore, NaOH will have the highest pH.

Final answer:

NaOH solution will have the highest pH.

Explanation:

The solution with the highest pH would be the one that is basic, as basic solutions have higher pH values. Among the given options, NaOH is a strong base and dissociates completely in water to produce OH- ions, which makes the solution highly basic.