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Natali5045456 [20]
3 years ago
11

A compound is 80.0% carbon and 20.0% hydrogen by mass. assume you have a 100.-g sample of this compound. the molar mass of the c

ompound was found to be 30.069 g/mol. what is the molecular formula?
Chemistry
1 answer:
ch4aika [34]3 years ago
4 0
Basis of the calculation: 100g
 
For Carbon: 
 Mass of carbon = (100 g)(0.80) = 80 g
  Number of moles of carbon = (80 g)(1 mole / 12g) = 20/3

For Hydrogen:
  Mass of hydrogen = (100 g)(0.20) = 20 g
     Number of moles of hydrogen = (20 g)(1 mole / 1 g) = 20

Translating the answer to the formula of the substance,
     C20/3H20

Dividing the answer,
    CH3

The molar mass of the empirical formula is:
    12 + 3 = 15 g/mol

Since, the molar mass given for the molecular formula is 30.069 g/mol, the molecular equation is,
    C2H6

ANSWER: C2H6

 
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Answer:

left to right across a period when it decreases and when it increases top to bottom in a group,

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2 years ago
The activation energy for the reaction NO2(g)+CO(g)⟶NO(g)+CO2(g) is Ea = 75 kJ/mol and the change in enthalpy for the reaction i
lapo4ka [179]

Answer:

= 100kJ

Explanation:

The reverse reaction's activation energy of a reaction is the activation energy of the forward reaction plus ΔH of the reaction:

Ea of forward reaction =75kJ

∆H = -175 kJ/mol

Ea of reverse reaction = 75 +(-175)

= 100kJ

Note that a reverse reaction is one which can proceed in both direction depending on the conditions.

6 0
3 years ago
3. How many grams of potassium permanganate will react with 36 grams of iron(III)<br>hydroxide? ​
Mrac [35]

7 would be correct because I divide it

8 0
3 years ago
A) ambos pertencem ao 5o período da Tabela Periódica;
Reptile [31]

Answer:

A

Explanation:

8 0
3 years ago
A) Find the gas speed of sulfur dioxide at 100.0 degrees Celsius? ______________
gtnhenbr [62]

a. 381.27 m/s

b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide

<h3>Further explanation</h3>

Given

T = 100 + 273 = 373 K

Required

a. the gas speedi

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  

Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol

\tt v=\sqrt{\dfrac{3\times 8.314\times 373}{0.064} }\\\\v=381.27~m/s

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 sulfur dioxide = 64

M₂ =  molar mass nitrogen triodide = 395

\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{395}{64} }=\dfrac{20}{8}=2.5

the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide

4 0
2 years ago
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