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andrew11 [14]
2 years ago
15

Determining the molecular formula of a hydrocarbon with a relative molecular mass of 54​

Chemistry
1 answer:
frez [133]2 years ago
8 0

Answer:

C₄H₆

Explanation:

A hydrocarbon is a molecule that only contains atoms of Carbon and Hydrogen. As you remember, the molar mass of Carbon is 12g/mol and the molar mass of Hydrogen is 1g/mol.

To solve this question we can find the amount of Carbons necessary that doesn't exceed the 54g/mol. That is:

54g/mol / 12g/mol = 4.5

As the molecular formula must be given in whole numbers, the amount of Carbons is 4. This 4 carbons have a mass of:

12g/mol*4 = 48g/mol

And the hydrogens necessaries to suply the molecular mass are:

54 - 48 = 6 hydrogens

That means molecular formula is:

<h3>C₄H₆</h3>
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You have 15.42g of C2H6. How many moles of H2O can be made?
Amiraneli [1.4K]

<u>Answer:</u> The moles of water produced are 1.54 moles.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of ethane = 15.42 g

Molar mass of ethane = 30.07 g/mol

Putting values in above equation, we get:

\text{Moles of ethane}=\frac{15.42g}{30.07g/mol}=0.513mol

The chemical equation for the combustion of ethane follows:

2C_2H_6+5O_2\rightarrow 4CO2+6H_2O

By Stoichiometry of the reaction:

2 moles of ethane produces 6 moles of water

So, 0.513 moles of ethane will produce = \frac{6}{2}\times 0.513=1.54mol of water

Hence, the moles of water produced are 1.54 moles.

4 0
3 years ago
Which of the following sets of oxidation numbers is NOT correct? KMnO4 : K +2, Mn +6, O -2 NaCHO2 : Na +1, C +2, H +1, O -2 Zn(O
muminat

KMnO4 has the incorrect set of oxidation numbers. Whenever there is an alkali metal, it has an oxidation number of +1. As you can see, K is said to have an oxidation number of +2, so it is incorrect.


6 0
3 years ago
A student uses a calorimeter to determine the enthalpy of dissolving for ammonium nitrate. The student fills a calorimeter with
ale4655 [162]
Enthalpy change during the dissolution process = m c ΔT,

here, m = total mass = 475 + 125 = 600 g
c = <span>specific heat of water = 4.18 J/g °C
</span>ΔT = 7.8 - 24 = -16.2 oc (negative sign indicates that temp. has decreases)
<span>
Therefore, </span>Enthalpy change during the dissolution = 600 x 4.18 X (-16.2)
                                                                                 = -40630 kJ
(Negative sign indicates that process is endothermic in nature i.e. heat is taken by the system)

Thus, <span>enthalpy of dissolving of the ammonium nitrate is -40630 J/g</span>

7 0
3 years ago
A sheet of polyethylene 1.5-mm thick is being used as an oxygen barrier at a temperature of 600K. If the flux is 2.48 x 10-5 kg/
Eddi Din [679]

Complete Question

The complete question is shown on the first uploaded image  

Answer:

The concentration of  high-pressure side is  C_1 =   0.722 \ kg/m^3

Explanation:

From the question we are told that

    The thickness of the polyethylene is  d  =  1.5 \ mm = 0.0015 \ m

     The  temperature is  T  =  600 \   K

      The flux is  JA  =  2.48 *10^{-5} \  kg/m^2\cdot s

      The concentration on the low-pressure side is  C_2 =  0.5 \ kg/m^3

       The initial diffusivity  is  D_o  =  6.2 *10^{-4} \ m^2 /s

       The activation energy for  diffusion is   Q_d  =  41 \ kJ /mol  =  41*10^3 J /mol

Generally the diffusivity  of the oxygen at 600 K can be  mathematically evaluated  as

        D   = D_o * e^{- \frac{Q_d}{R * T  } }  

substituting values  

         D   = 6.2*10^{-4} * e^{- \frac{41 *10^3}{8.314 * 600  } }  

          D   = 1.671 *10^{-7} \ m^2 /s  

Generally the flux is mathematically represented as

          JA  =  D  *  \frac{C_1 -C_2}{d}

Where C_1 is the concentration of oxygen at the higher side

       So  

             C_1 =    d  * \frac{JA}{D}  + C_2

substituting values  

             C_1 =    0.0015   * \frac{2.48*10^{-5}}{1.671*10^{-7}}  + 0.5

              C_1 =   0.722 \ kg/m^3

 

6 0
3 years ago
Please help IM IN HURRY!!!
Vladimir [108]
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