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Sophie [7]
3 years ago
13

What is the empirical formula for Hg2(NO3)2

Chemistry
1 answer:
den301095 [7]3 years ago
8 0
<span>Answer: HgNO₃
</span><span />

<span>Explanation:
</span>
<span /><span /><span>
The empirical formula is the formula that shows the ratio of the atoms in its simplest form, this is using the smallest whole numbers.
</span><span />

<span>The empirical formula may or may not be the same molecular formula.
</span><span />

<span>In this case you are given the molecular formula Hg₂(NO₃)₂. Since, the ratio of the atoms of Hg, N, and O is 2: 2: 6, respectively, the same ratio is expressed if you divide by the greatest common factor (GCF).
</span><span>
</span><span>
</span><span>The GCF of 2, 2, and 6 is 2. So, the ratios can be simplified to 1:1:3, meaning 1 mol of Hg, 1 mol of N, and 3 mol of O or HgNO₃.</span>
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4 0
3 years ago
How do I find molar mass​
sergij07 [2.7K]

Answer:The molar mass is the mass of a given chemical element or chemical compound (g) divided by the amount of substance (mol).

The molar mass of a compound can be calculated by adding the standard atomic masses (in g/mol) of the constituent atoms.

Explanation:

7 0
2 years ago
A solution contains 0.60 mg/ml mn2+. what minimum mass of kio4 must me added to 5.00 ml of the solution in order to completely o
azamat
The given solution of Mn²⁺ is 0.60 mg/mL.
Hence mass of Mn²⁺ in 5 mL of solution = 0.60 mg/mL x 5 mL = 3 mg

Molar mass of Mn = 54.9 g/mol
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The balanced equation for the reaction is,
2Mn²⁺ + 5KIO₄ + 3H₂O → 2MnO₄⁻ + 5KIO₃ + 6H⁺

The stoichiometric ratio between Mn²⁺ and KIO₄ is 2 : 5

Hence, moles of KIO₄ reacted = 5.46 x 10⁻⁵ mol x (5 / 2)
                                                 = 13.65 x 10⁻⁵ mol
Molar mass of KIO₄ = 230 g/mol

Hence needed mass of KIO₄ = 13.65 x 10⁻⁵ mol x 230 g/mol
                                               = 0.031395 g
                                               = 31.395 mg
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5 0
3 years ago
Give two examples in your everyday experience where diffusion occurs. Can you think of a situation of where this might be harmfu
Vesnalui [34]
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8 0
3 years ago
A flash distillation chamber operating at 101.3 kpa is separating an ethanol water mixture the feed mixture contains z weight et
Alex73 [517]

Answer:

The answer is [\frac{LX_1}{46} + \frac{L(1-x)}{18}]\times0.454 mol/hr

[\frac{Vy_1}{46}+\frac{V(1-y)}{18}]\times0.454mol/hr

Explanation:

For flash distillation

F = V+L

\frac{V}{F} + \frac{L}{F} = 1

\frac{F}{V} -\frac{L}{V} = 1

Fz = Vy+Lx

Y = \frac{F}{V}\times Z - \frac{L}{V}\times X                  let, \frac{V}{F} = F

y = \frac{Z}{F} -[ \frac{1}{F} -1]\times X

Highlighted reading

F = 299;  \frac{V}{F} = 0.85 ; z = 0.36

y = \frac{0.36}{0.85} - (-0.15)\times X

 = 0.423 + 0.15x ------------(i)

y^{*} = -43.99713x^{6} + 148.27274x^{5} - 195.46x^{4}+127.99x^{3}-43.3x^{2}+ 7.469x^{}+ 0.02011

At equilibrium, y^{*} = y

0.423+0.15x^{} = y^{*}

-43.99713x^{6}+ 148.27274x^{5} - 195.46x^{4}+127.99x^{3}-43.3x^{2}+ 7.319x^{}-0.403

F(x) for Newton's Law

Let x_{0} = 0

     x_{1}     = \frac{0-[{-0.403}]}{7.319}

             = 0.055

     x_{2}      = \frac{{0.055}-{f(0.055)} {{{{{{{}}}}}}}}{f^{'} (0.055)}

             = \frac{{(0.055)}-(-0.11)}{3.59}

             = 0.085

    x^{3}    = \frac{{0.085}-(0.024)}{2.289}

           = 0.095

   x^{4}     = \frac{{0.095}-(-0.0353)}{-1.410}

            = 0.07

From This x and y are found from equation (i) and L and V are obtained from \frac{V}{F}  and F values

[\frac{LX_1}{46} + \frac{L(1-x)}{18}]\times0.454 mol/hr

[\frac{Vy_1}{46}+\frac{V(1-y)}{18}]\times0.454mol/hr

   

8 0
3 years ago
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