All categories

3 years ago

Based on the reactivities of the elements involved, which reaction will form products that are more stable than the reactants?

Chemistry

1 answer:

vladimir2022 [97]3 years ago

5 0

Answer : The correct option is, (B)

Explanation :

According to the reactivity series of the elements, a chemical reaction occurs when the most reactive element displace the least reactive element and form more stable products as compared to the reactants.

Option A reaction : $2AlBr_3+3Zn\rightarrow \text{No reaction}$

In this, no reaction will occur because zinc is less reactive metal as compared to aluminium. So, zinc will not displace the aluminium.

Option B reaction : $CaBr_2+2Na\rightarrow 2NaBr+Ca$

In this, the reaction will occur because sodium is more reactive metal as compared to calcium. So, sodium will displace the calcium easily.

Option C reaction : $MgBr_2+H_2\rightarrow \text{No reaction}$

In this, no reaction will occur because hydrogen is less reactive metal as compared to magnesium. So, hydrogen will not displace the magnesium.

Option D reaction : $BaBr_2+Ca\rightarrow \text{No reaction}$

In this, no reaction will occur because calcium is less reactive metal as compared to barium. So, calcium will not displace the barium.

Option E reaction : $2LiBr+Ba\rightarrow \text{No reaction}$

In this, no reaction will occur because barium is less reactive metal as compared to lithium. So, barium will not displace the lithium.

Hence, the option B reaction will form products that are more stable than the reactants.

The reactivity series of the elements are shown below.

You might be interested in

pickupchik [31]

Answer:

higher infant mortality

Explanation:

3 0

3 years ago

What changes in space result from the death of massive stars?

lana [24]

Whenever the fuel is being used up, a star explodes and the energy leakage from a star's core ceases.

Explanation:

The dying star expands in the "Red Giant," before even the inevitable collapse starts, due to nuclear reactions just outside of the core.

It becomes a white dwarf star when the star has almost the same density as the Sun. If it's much larger, a supernova explosion could take place and leave a neutron star away. However, if it is very large–at least three times the Sun's mass–the crumbling core of the star, nothing will ever stop it from crumbling. The star is imploding into a black hole, an endless gravitational loop in space.

6 0

3 years ago

WILL GIVE BRAINLIEST - A solution has a [H3O+] of 1 × 10−3 M. What is the [OH−] of the solution?

Hitman42 [59]

Answer:

The answer is 1 x 10 - 11 M. :))

4 0

3 years ago

Valence electrons (electrons in the outermost shell – available for bonding)

Anastaziya [24]

Electrons and newtrons is that what ur asking for

8 0

3 years ago

Give the name and formula of the compound formed from the following elements: a) Sodium and nitrogen b) Oxygen and strontium c)

elena-s [515]

Answer: a) Sodium and nitrogen : $NaN_3$ : sodium nitride.

b) Oxygen and strontium : $SrO$ : strontium oxide.

c) Aluminum and chlorine : $AlCl_3$ : aluminium chloride.

d) Cesium and bromine : $CsBr$ : cesium bromide.

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

a) Sodium and nitrogen : Here sodium is having an oxidation state of +1 called as $Na^{+}$ cation and nitrogen forms an anion $N^{3-}$ with oxidation state of -3. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $NaN_3$ named as sodium nitride.

b) Oxygen and strontium : Here strontium is having an oxidation state of +2 called as $Sr^{+}$ cation and oxygen forms an anion $O^{2-}$ with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $SrO$ named as strontium oxide.

c) Aluminum and chlorine : Here aluminium is having an oxidation state of +3 called as $Al^{3+}$ cation and chlorine forms an anion $Cl^{-}$ with oxidation state of -1. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $AlCl_3$ named as aluminium chloride.

d) Cesium and bromine : Here cesium is having an oxidation state of +1 called as $Cs^{+}$ cation and bromine forms an anion $Br^{-}$ with oxidation state of -1. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $CsBr$ named as cesium bromide.

7 0

3 years ago