Answer: Plot [NO] versus time, and calculate the slope of a line drawn tangent to the point of interest on the graph.
Explanation: The instantaneous rate of reaction is the change in concentration over an infinitesimally small time interval.
That is, (ΔC/Δt) as Δt --> 0 or dC/dt
Instantaneous rate can be obtained from experimental data by first graphing the concentration of a system as function of time, and then finding the slope of the tangent line at a specific point which corresponds to a particular time of interest.
Alternatively, one can measure the change in concentration over a very small time period two or more times to get an average rate close to that of the instantaneous rate. The reaction rate for that time is determined from the slope of the tangent lines still.
Below the ph of 7.0 is acidic. Anything above is basic and the ph of 7.0 is neutral
The correct answer for the question that is being presented above is this one: "A. A sample of 0.500 mole of O2 is added to the 4.80 g of O2 in the container." A sample containing 4.80 g of O2 gas has an initial volume of 15.0 L. The final volume, in liters, when each of the following changes occurs in the quantity of the gas at a constant pressure and temperature is that <span>A sample of 0.500 mole of O2 is added to the 4.80 g of O2 in the container.</span>
Geologists look for texture, color, and mineral composition. Since they can be found anywhere such as oceans, deserts, and mountains. <span />
Answer:
Final volume=V₂ = 216.3 mL
Explanation:
Given data:
Initial volume = 120.0 mL
Initial temperature = -12.3 °C (-12.3 +273 = 260.7 K)
Final volume = ?
Final temperature = 197.0 °C (197+273 = 470 K)
Solution:
We will apply Charles Law to solve the problem.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 120 mL × 470 K /260.7K
V₂ = 56400 mL.K /260.7K
V₂ = 216.3 mL
