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strojnjashka [21]
3 years ago
5

HgC2H6

Chemistry
1 answer:
Fittoniya [83]3 years ago
6 0

Answer:

The number of molecules= 1.33 × 10∧22 molecules

percentage of mercury = 87%

Explanation:

Given data:

mass of dimethyl mercury = 5.10 g

number molecules of dimethyl mercury in 5.10 g = ?

percentage of mercury = ?

Solution:

First of all we will calculate the molar mass of dimethyl mercury.

molar mass of HgC2H6 = 1×200.6 + 2×12 + 6×1 = 230.6 g/mol

we know that,

230.6 g of HgC2H6 = 1 mol = 6.02 × 10∧23 molecules.

so

For the 5.10 g of sample:

5.10 g/230 g/mol = 0.022 moles of HgC2H6

now we will multiply these number of moles with Avogadro number to get number of molecules in 5.10 g of sample.

0.022 × 6.02 × 10∧23 molecules = 0.133  × 10∧23 molecules or

1.33 × 10∧22 molecules.

Percentage of mercury:

Formula:

percentage = (atomic number of Hg × total number of atoms of Hg/ molar mass of HgC2H6) × 100

% age of Hg = (200.6 g/mol× 1/ 230.6 g/mol) × 100

%age of Hg = 0.869 × 100

%age of Hg = 86.99 % or 87 %

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Doss [256]

Answer:

C6H14O3F

Explanation:

The first step is to divide each compound by its molecular weight

Carbon

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Hydrogen

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Oxygen

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Phosphorous

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The next step is to divide by the lowes value

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7.67/0.542

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3 years ago
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A) Find the gas speed of ammonia at 25.0 degrees Celsius. ______________
Cloud [144]

a. 661.23 m/s

b.  the rate of effusion of Ammonia = 4.5 faster than Silicon tetra bromide

<h3> Further explanation </h3>

Given

T = 25 + 273 = 298 K

Required

a. the gas speed

b. The rate of effusion comparison

Solution

a.

Average velocities of gases can be expressed as root-mean-square averages. (V rms)  

\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}

R = gas constant, T = temperature, Mm = molar mass of the gas particles  

From the question  

R = 8,314 J / mol K  

T = temperature  

Mm = molar mass, kg / mol  

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b. the effusion rates of two gases = the square root of the inverse of their molar masses:  

\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }

M₁ = molar mass Ammonia NH₃= 17

M₂ =  molar mass Silicon tetra bromide SiBr₄= 348

\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{348}{17} }=4.5

the rate of effusion of Ammonia = 4.5 faster than Silicon tetra bromide

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Answer: I think its C or B

Explanation: Hope this was helpful....

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