1 Atomic mass unit is the mass of an atom or it caliberates mass on atomic scale. It is also expressed in dalton denoted by Da whereas Atomic mass unit is denoted by amu.
1 amu can be expressed in grams as follows:
For converting in grams,
Thus, mass of Te is 204.16 * 10^-2^4 g
The mole fraction of methanol in the mixture is 0.444
We'll begin by calculating the number of mole of water.
- Molar mass of water = 18 g/mol
Mole = mass / molar mass
Mole of water = 45 / 18
Mole of water = 2.5 moles
Finally, we shall determine the mole fraction of methanol.
- Mole of water = 2.5 moles
- Mole of methanol = 2 moles
- Total mole = 2 + 2.5 = 4.5 moles
Mole fraction of methanol =?
Mole fraction = mole / total mole
Mole fraction of methanol = 2 / 4.5
Mole fraction of methanol = 0.444
Thus, the mole fraction of methanol is 0.444
Learn more about mole fraction:
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Answer:
Q = 0.061 = Kc
Explanation:
Step 1: Data given
Temperature = 500 °C
Kc=0.061
1.14 mol/L N2
5.52 mol/L H2
3.42 mol/L NH3
Step 2: Calculate Q
Q=[products]/[reactants]=[NH3]²/ [N2][H2]³
If Qc=Kc then the reaction is at equilibrium.
If Qc<Kc then the reaction will shift right to reach equilibrium.
If Qc>Kc then the reaction will shift left to reach equilibrium.
Q = (3.42)² / (1.14 * 5.52³)
Q = 11.6964/191.744
Q = 0.061
Q = Kc the reaction is at equilibrium.
Answer:
Products are favored.
Explanation:
The acid-base reaction of CH₃COOH (acid) with NH₃ (base) produce:
CH₃COOH + NH₃ ⇄ CH₃COO⁻ + NH₄⁺ Kr = ?
It is possible to know Kr of the reaction by the sum of acidic dissociations of the half-reactions. That is:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ Ka = 1.8x10⁻⁵
NH₃ + H⁺ ⇄ NH₄⁺ 1/Ka = 1/ 5.6x10⁻¹⁰ = 1.8x10⁹
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CH₃COOH + NH₃ ⇄ CH₃COO⁻ + NH₄⁺ Kr = 1.8x10⁻⁵×1.8x10⁹ = <em>3.2x10⁴</em>
<em> </em>
As Kr is defined as:
Kr = [CH₃COO⁻] [NH₄⁺] / [CH₃COOH] [NH₃]
And Kr is > 1
[CH₃COO⁻] [NH₄⁺] > [CH₃COOH] [NH₃],
showing <em>products are favored</em>.
