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Marrrta [24]
3 years ago
15

How many moles of methane were in the erlenmeyer flask?

Chemistry
2 answers:
DanielleElmas [232]3 years ago
5 0

Since the question is incomplete, I can help you with a detailed explanation on how to calculate the number of moles of mehate in an erlenmeyer flask.


There are two possible scenaries for this question.


1) Scenary I. Known mass.


Whether you are given the mass or other information that permits to calculate the mass of methane, you can work with the molecular formula to calculate the number of moles, in this way:


i) Methane chemical formula: CH₄


ii) Molar masss: sum the mass of each element in the formula, using the atomic mass and number of atoms of each.


Molar mass = 12.011 g/mol + 4× 1.008 g/ mol = 16.043 g/mol


iii) Number of moles


Formula: number of moles = mass in grams / molar mass = mass in grams / 16.043 g/mol.


That is it. Use the last formula to calculate the number of moles of methane.


2) Scenary 2. P, T,and V known.


You do not know the mass but the conditions: temperature (T), volume (V), and pressure (P).


In this case you use the ideal gas equation:


PV = nRT, from which you clear for the number of moles (n):


n = PV/(RT).


Plug in the data of P, V and T in coherent unit system.


R is the universal law of gases, which you must also use with the appropiate units, depending on the units of P, V, and T used.


For example, R = 0.0821 atm×liter / (K×mol); and R = 8.314 KPa×liter/(mol×K).

Savatey [412]3 years ago
4 0
Methane Volume : O.O
Methane Mass: 0.100g
Molar Mass of Methane : 16.04 g/mol
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Explanation:

To solve this question we can use the equation:

ΔG = -RTlnK

<em>Where ΔG is Gibbs free energy = 12.86kJ/mol</em>

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<em>T is absolute temperature = 298K</em>

<em>And K is equilibrium constant.</em>

Replacing:

12.86kJ/mol = -8.314x10⁻³kJ/molK*298K lnK

5.19 = lnK

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Calculate the number of moles of magnesium, chlorine, and oxygen atoms in 4.90 moles of magnesium perchlorate, Mg(ClO4)2.]\
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Answer:

 4.90  moles of  Mg(ClO_4)_2   will produce  (9.8) moles of  Cl^{-} ,

                                                    (4.90) moles of Mg^{2+} and

                                                    (39.2) moles of  O^{2-}

Explanation:

From the question we are told that

  The number of moles of  is  n =  4.90  \ mols

The formation reaction of Mg(ClO_4)_2  is

             Mg^{2+} + 2 Cl^{-} + 8O^{2+} \to Mg(ClO_4)_2  

From the reaction we see that

   1 mole of  Mg(ClO_4)_2  is formed by 2 moles of Cl^{-} 1 mole of  Mg^{2+} and 4  O^{2-}

 This implies that

   4.90  moles of  Mg(ClO_4)_2   will produce  (2 * 4.90) moles of  Cl^{-} ,

                                                    (1 * 4.90) moles of Mg^{2+} and

                                                    (8 * 4.90) moles of  O^{2-}

So

  4.90  moles of  Mg(ClO_4)_2   will produce  (9.8) moles of  Cl^{-} ,

                                                    (4.90) moles of Mg^{2+} and

                                                    (39.2) moles of  O^{2-}

                           

4 0
3 years ago
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