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Answer: 1.037M
Explanation:
Since the rate constant unit is per seconds, therefore it is a first order reaction.
First order reaction equation is given as
InA= -kt +InAo
Where,Ao is the initial concentration of reactant =0.600M
A is the concentration of reactant at a specifies time t=3×60=180s
and k is the rate constant
InA = -6.50×10^-3 ×180 +In(0.6)
InA = -1.17 + 0.5108
InA= -1.680
A = e-1.680
A= 1.037M
Therefore the concentration after 3minutes is 1.037M
Answer:
number of carbon-carbon single (C - C) bonds: 1
number of carbon-hydrogen single (C H) bonds: 5
number of nitrogen-hydrogen sing le (N H) bonds:2
number of lone pairs: 1
Explanation:
Ethanamine is a colourless gas having a strong 'ammonia- like' odour. It contains the -NH2 group which makes it an amine. It contains one carbon-carbon bond, five carbon-hydrogen bonds and two nitrogen-hydrogen bonds.
Nitrogen, being sp3 hybridized in the compound has a lone pair of electrons localized on one of the sp3 hybridized orbitals of nitrogen while one sp3 hybridized orbital of nitrogen is used to form a carbon-nitrogen bond. The other two sp3 hybridized orbitals on nitrogen are used to form the two nitrogen-hydrogen bonds.
The Henderson-Hasselbalch equation can be used to determine the pH of the buffer from the pKa value. The pH of the buffer will be 4.75.
<h3>What is the Henderson-Hasselbalch equation?</h3>
Henderson-Hasselbalch equation is used to determine the value of pH of the buffer with the help of the acid disassociation constant.
Given,
Acid disassociation constant (ka) = 1. 8 10⁻⁵
Concentration of NaOH = 2.0 M
Concentration of CH₃COOH = 2.0 M
pKa value is calculated as,
pKa = -log Ka
pKa = - log (1. 8 x 10⁻⁵)
Substituting the value of pKa in the Henderson-Hasselbalch equation as
pH = - log (1. 8 x 10⁻⁵) + log [2.0] ÷ [2.0]
pH = - log (1. 8 x 10⁻⁵) + log [1]
= 4.745 + 0
= 4.75
Therefore, 4.75 is the pH of the buffer.
Learn more about the Henderson-Hasselbalch equation here:
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Answer:
The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Explanation:
Given;
CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol
From the combustion reaction above, it can be observed that;
1 mole of methane (CH₄) released 890 kilojoules of energy.
Now, we convert 59.7 grams of methane to moles
CH₄ = 12 + (1x4) = 16 g/mol
59.7 g of CH₄ 
1 mole of methane (CH₄) released 890 kilojoules of energy
3.73125 moles of methane (CH₄) will release ?
= 3.73125 moles x -890 kJ/mol
= -3320.81 kJ
Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
