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topjm [15]
3 years ago
7

Aqueous magnesium chloride is added to aqueous silver nitrate. A white precipitate forms. Write the chemical equation for this r

eaction.
Chemistry
2 answers:
Sindrei [870]3 years ago
7 0

Answer:

The symbol for magnesium chloride is MgCl2 and silver nitrate is Ag(NO3)2 so the equation will be :

MgCl2 + Ag(NO3)2 → AgCl2 + Mg(NO3)2

Butoxors [25]3 years ago
4 0

Answer:

1. Molecular equation

BaCl2(aq) + 2AgNO3(aq) –> 2AgCl(s) + Ba(NO3)2 (aq)

2. Complete Ionic equation

Ba²⁺(aq) + 2Cl¯(aq) + 2Ag⁺(aq) + 2NO3¯ (aq) —> 2AgCl(s) + Ba²⁺(aq) + 2NO3¯(aq)

3. Net ionic equation

Cl¯(aq) + Ag⁺(aq) —> AgCl(s)

Explanation:

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Answer is 0.289nm.

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wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}

For calculating edge length,

a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}

For calculating M_{ave}, we use the formula

M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}

Similarly for calculating (\rho)_{ave}, we use the formula

\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}

From the periodic table, masses of the two elements can be written

M_{Fe}= 55.85g/mol

M_{V}=50.941g/mol

Specific density of both the elements are

(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3

Putting  M_{ave} and \rho_{ave} formula's in edge length formula, we get

a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}}  \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}}  \right )}  \right ]^{1/3}

a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}}  \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}}  \right )}  \right ]^{1/3}

By calculating, we get

a=2.89\times10^{-8}cm=0.289nm

7 0
2 years ago
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