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tankabanditka [31]
3 years ago
8

Write a balanced equilibrium equation for the dissolution of nai in water. include phases.

Chemistry
2 answers:
Lorico [155]3 years ago
4 0

Explanation:

The compound sodium iodide (NaI) is an ionic compound as there is formation of chemical bond by transfer of electrons from sodium to iodine.

Therefore, it is a polar molecule and we know that water is also a polar solvent. Hence, like dissolves like as a result, sodium iodide will dissociate into water.

The reaction equation will be as follows.

           NaI(s) + H_{2}O \rightarrow Na^{+}(aq) + I^{-}(aq)

Bess [88]3 years ago
3 0
Dissolution means to make the compound apart, So when we have ionic compounds like NaI which has metal and non-metal ions, It separates into parts of positive ions and negative ions. After we separate this compound apart we will put the charge of each on above its symbol and then start to balance the equation of the dissolution.

So the dissolution equation of NaI is:
NaI(s) → Na^+(s)  +   I^-(Aqu) 
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Mrrafil [7]

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2 years ago
A 32.2 g iron rod, initially at 21.9 C, is submerged into an unknown mass of water at 63.5 C. in an insulated container. The fin
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Answer : The mass of the water in two significant figures is, 3.0\times 10^1g

Explanation :

In this case the heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of iron metal = 0.45J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of iron metal = 32.3 g

m_2 = mass of water = ?

T_f = final temperature of mixture = 59.2^oC

T_1 = initial temperature of iron metal = 21.9^oC

T_2 = initial temperature of water = 63.5^oC

Now put all the given values in the above formula, we get

32.3g\times 0.45J/g^oC\times (59.2-21.9)^oC=-m_2\times 4.18J/g^oC\times (59.2-63.5)^oC

m_2=30.16g\approx 3.0\times 10^1g

Therefore, the mass of the water in two significant figures is, 3.0\times 10^1g

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