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3 years ago

Does the H₂SO4 solution contain ions? O yes O no

Chemistry

1 answer:

Artist 52 [7]3 years ago

8 0

$\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the concept of Ions .

If we see the reaction.

H2SO4 ==> 2H+ + (SO4)2-

Here you can see, H+ ( Hydrogen ions) and (SO4) 2- ( sulphate ions) are the ions when the H2SO4 is dissolved in water.

You might be interested in

What is the hybridization of carbon in nco-?

Svet_ta [14]

First let us see what kind of bonds are formed in the compound. By drawing the structure, we see that the kind of bonds are:

N =- triple bond -= C – O

So there is only single bond between C and O therefore the hybridization of C is sp.

4 0

3 years ago

Read 2 more answers

Consider the insoluble compound nickel(II) hydroxide , Ni(OH)2 . The nickel ion also forms a complex with cyanide ions . Write a

natali 33 [55]

Answer: Equilibrium constant for this reaction is $2.8 \times 10^{15}$ .

Explanation:

Chemical reaction equation for the formation of nickel cyanide complex is as follows.

$Ni(OH)_{2}(s) + 4CN^{-}(aq) \rightleftharpoons [Ni(CN)_{4}^{2-}](aq) + 2OH^{-}(aq)$

We know that,

K = $K_{f} \times K_{sp}$

We are given that, $K_{f} = 1.0 \times 10^{31}$

and, $K_{sp} = 2.8 \times 10^{-16}$

Hence, we will calculate the value of K as follows.

K = $K_{f} \times K_{sp}$

K = $(1.0 \times 10^{31}) \times (2.8 \times 10^{-16})$

= $2.8 \times 10^{15}$

Thus, we can conclude that equilibrium constant for this reaction is $2.8 \times 10^{15}$ .

4 0

3 years ago

Draw the Lewis structure for the compound with the formula COCl2COCl2. Use lines to show bonding electrons.

stira [4]

Answer : The Lewis-dot structure of $COCl_2$ is shown below.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, $COCl_2$

As we know that carbon has '4' valence electrons, chlorine has '7' valence electron and oxygen has '6' valence electrons.

Therefore, the total number of valence electrons in $COCl_2$ = 1(4) + 2(7) + 1(6) = 24

According to Lewis-dot structure, there are 8 number of bonding electrons and 16 number of non-bonding electrons.

6 0

4 years ago

Common table salt is sodium chloride. when table salt is dissolved in water, the solution conducts electricity because

enot [183]

In commonly table salt is sodium chloride. When table salt is dissolved in water, the solution conducts electricity because sodium chloride is made up of ions, it is called an ionic substance.

4 0

4 years ago

During studies of the following reaction (i), a chemical engineer measured a less-than-expected yield of N2 and discovered that

bonufazy [111]

Answer:

Maximum expected yield = 87.2%

Explanation:

Equations of reactions:

Main reaction: N₂O₄(l) + 2N₂H₄(l) ---> 3N₂(g) + 4H₂O(g)

Side reaction: 2N₂O₄(l) + N₂H₄(l) ----> 6NO(g) + 2H₂O(g)

Molar mass of N₂O₄ = 92 g/mol; molar mass of N₂H₄ = 32 g/mol; molar mass of N₂ = 28 g/mol; molar mass of of NO = 30 g/mol; molar mass of water = 18 g/mol

In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂.

101.1 g of N₂O₄ will react with 2 * 32 * 101.1 / 92 g of N₂H₄ = 70.33 g of N₂H₄

N₂O₄ is the limiting reactant

101.1 g of N₂O₄ will react to produce 3 * 14 * 101.1 / 92 g of N₂ = 46.15 g of N₂

In the side reaction, (6 * 30 g) of NO is produced from (2 * 92 g) of N₂O₄ and 32 g of N₂H₄

12.7 g of N₂O₄ will be produced from ( 2 * 92 * 12.7/180 g) of N₂O₄ and (32 * 12.7/180) g of N₂H₄ to produce

mass of N₂O₄ used = 12.98 g

mass of N₂H₄ used = 2.26 g

mass of N₂O₄ left for main reaction = 101.1 - 12.98 = 88.12 g

mass of N₂H₄ left for main reaction = 101.1 - 2.26 = 98.84 g

In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂

88.12 g of N₂O₄ will react with 2 * 32 * 88.12 / 92 g of N₂H₄ = 61.30 g of N₂H₄

N₂O₄ is the limiting reactant.

88.12 g of N₂O₄ will to react produce 3 * 14 * 88.12 / 92 g of N₂ = 40.23 g of N₂

Percentage yield = (theoretical yield/actual yield) * 100%

Percentage yield = (40.23/46.15) * 100% = 87.2%

Therefore, maximum expected yield = 87.2%

4 0

3 years ago