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3 years ago

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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You wish to make 250. mL of 0.20 M KCl from a stock solution of 1.5 M KCl and DI water.

Nitella [24]

The question requires us to use the dilution formula $M_iV_i=M_fV_f$ , where $M_i$ and $V_i$ are the stock concentration and volume respectively, then $M_f$ and $V_f$ are the dilute concentration and volume respectively.

a. $C_s_t_o_c_k$ = 1.5 M KCI, $C_d_i_l_u_t_e$ =0.20M KCl, $V_d_i_l_u_t_e$ =250ml KCl

b. $M_iV_i=M_fV_f\\\implies V_i= \frac{M_fV_f}{M_i} = \frac{0.20M \times 250ml}{1.5M} = 33.3\ ml$

To prepare the solution 33.3ml of 1.5M KCl is diluted to a total final volume of 250ml.

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4 years ago

Atomic orbitals developed using quantum mechanics describe regions of space in which one is most likely to find an electron. giv

valina [46]

Answer:

Option A is correct.

Atomic orbitals developed using quantum mechanics describe regions of space in which one is most likely to find an electron

Explanation:

Atomic orbitals developed using quantum mechanics make use of quantum numbers.

There are four different quantum numbers that all work to give the region of space where a particular electron has the highest probability of being located.

The four quantum numbers that describes an electron's most likely location in an atom include

1) Principal quantum number, denoted by letter n. This quantum number gives the shell that an electron in an atom belongs to. It can take on natural number values from 1 (for the shell closest to the nucleus) through 2, 3, 4.... till rhe outermost shell.

2) Azimuthal/Angular Momentum quantum number, denoted by l. This quantum number describes the subshell or orbital within a shell that the electron belongs to in an atom.

It can take on values that can range from 0 to (n-1). These are the spdf orbitals with s-orbital having l-quantum number of 0, p-orbital with l-quantum number of 1 etc.

3) Magnetic quantum number, denoted by letter m. This describes the sub-orbital that the electron belongs to. It's values for electrons in a particular orbital vary from -l through 0 to +l.

E.g. orbital with l = 1 has electrons whose magnetic quantum number vary from -1, 0, +1.

orbital with l = 2 has electrons whose magnetic quantum number vary from -2, -1, 0, +1, +2.

4) Spin quantum number, denoted by letter s.

This describes the orientation of the electron's spin. Whether clockwise or anti-clockwise in it's sub-orbital. It can take on only values of (+1/2) or (-1/2).

So, these four quantum numbers, numbers that were made known because of quantum mechanics, show that atomic orbitals developed using quantum mechanics describe regions of space in which one is most likely to find an electron in an atom.

Hope this Helps!!!

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3 years ago