The answer is a. one. There are three phosphate bonds in the ATP but only the first bond is responsible for the high energy. When this bond break, ATP will change to ADP.
The number of moles of gas lost is 0.0213 mol. It can be solved with the help of Ideal gas law.
<h3>What is Ideal law ?</h3>
According to this law, "the volume of a given amount of gas is directly proportional to the number on moles of gas, directly proportional to the temperature and inversely proportional to the pressure. i.e.
PV = nRT.
Where,
- p = pressure
- V = volume (1.75 L = 1.75 x 10⁻³ m³)
- T = absolute temperature
- n = number of moles
- R = gas constant, 8.314 J*(mol-K)
Therefore, the number of moles is
n = PV / RT
State 1 :
- T₁ = (25⁰ C = 25+273 = 298 K)
- p₁ = 225 kPa = 225 x 10³ N/m²
State 2 :
- T₂ = 10 C = 283 K
- p₂ = 185 kPa = 185 x 10³ N/m²
The loss in moles of gas from state 1 to state 2 is
Δn = V/R (P₁/T₁ - P₂/T₂ )
V/R = (1.75 x 10⁻³ m³)/(8.314 (N-m)/(mol-K) = 2.1049 x 10⁻⁴ (mol-m²-K)/N
p₁/T₁ = (225 x 10³)/298 = 755.0336 N/(m²-K)
p₂/T₂ = (185 x 10³)/283 = 653.7102 N/(m²-K)
Therefore,
Δn = (2.1049 x 10⁻⁴ (mol-m²-K)/N)*(755.0336 - 653.7102 N/(m²-K))
= 0.0213 mol
Hence, The number of moles of gas lost is 0.0213 mol.
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Answer:
What are the advantages of titration?
Titrimetric analysis commonly referred to as volumetric analysis offers distinct advantages over cumbersome gravimetric methods:
Speed of analysis.
Instantaneous completion of reactions.
Greater accuracy due to minimization of material loss involved in decanting, filtration, precipitation or similar operations.
Explanation:
Disadvantages
It is a destructive method often using up relatively large quantities of the substance being analysed.
It requires reactions to occur in a liquid phase, often the chemistry of interest will make this inappropriate.
It can produce significant amounts of chemical waste which has to be disposed of.
It has limited accuracy.
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To determine the amount of 6.0 M H2SO4 needed for the preparation, equate the number of moles of the 6.0 M and 2.5 M H2SO4 solution. This is done as follows
M1 x V1 = M2 x V2
Substituting the known variables,
(6.0 M) x V1 = (2.5 M) x (4.8 L)
Solving for V1 gives an answer of V1 = 2 L. Thus, to prepare the needed solution, dilute 2 L of 6.0 M H2SO4 solution with water until the volume reach 4.8 L.
Critical pressure is the pressure of a gas or vapor in its critical state and critical point is a point on a phase diagram at which both the liquid and gas phases of a substance have the same density, and are therefore indistinguishable.
