An exergonic reaction proceeds with net release of free energy.
An endergonic reaction absorbs free energy
At constant temperature, if the pressure is compressed to the given value, the volume of the nitrogen gas increases to 23.2L.
<h3>What is Boyle's law?</h3>
Boyle's law simply states that "the volume of any given quantity of gas is inversely proportional to its pressure as long as temperature remains constant.
Boyle's law is expressed as;
P₁V₁ = P₂V₂
Where P₁ is Initial Pressure, V₁ is Initial volume, P₂ is Final Pressure and V₂ is Final volume.
Given that;
- Initial volume of the gas V₁ = 22.5L
- Initial pressure of the gas P₁ = 0.98atm
- Final pressure of the gas P₂ = 0.95atm
- Final volume of the gas V₂ = ?
P₁V₁ = P₂V₂
V₂ = P₁V₁ / P₂
V₂ = (0.98atm × 22.5L) / 0.95atm
V₂ = 22.05Latm / 0.95atm
V₂ = 23.2L
Therefore, at constant temperature, if the pressure is compressed to the given value, the volume of the nitrogen gas increases to 23.2L.
Learn more about Boyle's law here: brainly.com/question/1437490
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Hey there!
D = m / V
13.6 = 76.2 / V
V = 76.2 / 13.6
V = 5.602 mL
Answer:
It will be reported too low.
Explanation:
To measure the specific heat of the metal (s), the calorimeter may be used. In it, the metal will exchange heat with the water, and they will reach thermal equilibrium. Because it can be considered an isolated system (there're aren't dissipations) the total amount of heat (lost by metal + gained by water) must be 0.
Qmetal + Qwater = 0
Qmetal = -Qwater
The heat is the mass multiplied by the specific heat multiplied by the temperature change. If c is the specific heat of the water:
m_metal*s*ΔT_metal = - m_water *c*ΔT_water
s = -m_water *c*ΔT_water / m_metal*ΔT_metal
So, if m_water is now less than it was supposed to be, s will be reported too low, because they are directly proportional.
We are given that 1 teaspoon is equivalent to 5 mL,
therefore 0.75 teaspoon is:
0.75 teaspoon * (5 mL / 1 teaspoon) = 3.75 mL
So the mass is density times volume:
mass = (12.5 mg/5 ml) * 3.75 mL
<span>mass = 9.375 mg</span>
