Which are the states that donot have a seashore

Answer:

The reaction of white phosphorus with aqueous NaOH gives phosphine along with another phosphorus containing compound. The reaction type; the oxidation states of phosphorus in phosphine and the other product are respectively.

Answer:

White phosphorus reacts with caustic soda

White phosphorus is an extremely reactive element. White phosphorus ignites spontaneously in chlorine and reacts violently with the halogens. white phosphorus reacts very quickly with oxygen in the air, it may not be found far away from sources of contamination. The fate of white phosphorus smoke is similar to the fate of reaction products of white phosphorus.

White phosphorus reacts with caustic soda to give phosphine and sodium hypo phosphate.

P4 + 3H2O + 3NaOH → 3NaH2PO2 + PH3

White phosphorus reacts mainly with oxygen in the water and may stay in water for a period of hours to days.

Explanation:

the passenger car tata indica was launched by tata group. ( I think it is correct answer I write in board exam)

Answer:

The passenger car Tata Indica was launched by Tata Group.

Explanation:

Change the voice:

If the given sentence is in simple past tense do the following changes.

• find out what is the object of the given sentence

• passive voice sentence according to the given instructions.
• if the given sentence is in simple past tense use helping verbs was or were according to the subject.
• use past participle form of the verb after helping verb and then write down subject.

Answer:

(a) conservative X liberalist ( conservative has many antonyms)

(b) encourage X discourage

Answer:

Answer:

The solubility of [tex]BaSO_4[/tex] is more than [tex]Al(OH)_3[/tex] as the solubility of  [tex]BaSO_4[/tex] is [tex]1.048\times 10^-^5[/tex] and that of  [tex]Al(OH)_3[/tex] is [tex]10^-^8[/tex]. The solubility of [tex]Al(OH)_3[/tex] will decrease in presence of [tex]0.10M[/tex] [tex]NaOH[/tex] due to the common ion effect. And the solubility of  [tex]Al(OH)_3[/tex] in presence of a common ion effect will be [tex]2.7\times 10^-^2^8[/tex].

Explanation:

Let Solubility [tex]BaSO_4[/tex] of  be "S"

Then,

[tex]BaSO_4\rightarrow Ba^+^2+SO4^-^2\\[/tex]

                  S           S

Ksp of BaSO₄⇒[tex][Ba^+^2][SO4^-^2][/tex]     (1)

By substituting the value of Ksp of [tex]BaSO_4[/tex] in equation (1) we get;

[tex]1.1\times 10^-^1^0=[S][S][/tex]

[tex]S^2=1.1\times 10^-^1^0[/tex]

[tex]S=1.048\times 10^-^5[/tex]                (2)

Let Solubility of [tex]Al(OH)_3[/tex] be "S'"

Then,

[tex]Al(OH)_3\rightarrow Al^+^3+3OH^-^1[/tex]

                   S'          3S'

Ksp of Al(OH)₃⇒[tex][Al^+^3[3OH^-^1][/tex]

[tex]2.7\times 10^-^3^1=[S'][3S']^3[/tex]

[tex]S'=10^-8[/tex]                         (3)

From equations (2) and (3) we can see that the solubility of [tex]BaSO_4[/tex] is more than [tex]Al(OH)_3[/tex].

The solubility of [tex]Al(OH)_3[/tex] will decrease in presence of 0.10M NaOH due to the common ion effect.

Let's understand this by the reaction.

[tex]NaOH\rightarrow Na^++OH^-[/tex]

[tex]0.10M[/tex]     [tex]0.10M[/tex]   [tex]0.10M[/tex]

Total solubility of [tex][OH^-]\rightarrow{[OH^- from Al(OH)_3]+0.10}[/tex]

Now,

[tex]Ksp=2.7\times 10^-^3^1=[S'][[OH⁻ from Al(OH)₃]+0.10]^3[/tex]

We know that [tex][OH^-] < < < < 0.10[/tex]

So, we can neglect it.

Then,

[tex]2.7\times 10^-^3^1=S'[0.10]^3[/tex]

[tex]S'=2.7\times 10^-^2^8[/tex]

Hence, the solubility of [tex]BaSO_4[/tex] is more than [tex]Al(OH)_3[/tex] as the solubility of [tex]BaSO_4[/tex]  is [tex]1.048\times 10^-^5[/tex] and that of  [tex]Al(OH)_3[/tex] is [tex]10^-^8[/tex]. The solubility of [tex]Al(OH)_3[/tex] will decrease in presence of [tex]0.10M[/tex] [tex]NaOH[/tex] due to the common ion effect. And the solubility of [tex]Al(OH)_3[/tex] in presence of a common ion effect will be [tex]2.7\times 10^-^2^8[/tex].