To determine the time it takes to completely vaporize the given amount of water, we first determine the total heat that is being absorbed from the process. To do this, we need information on the latent heat of vaporization of water. This heat is being absorbed by the process of phase change without any change in the temperature of the system. For water, it is equal to 40.8 kJ / mol.
Total heat = 40.8 kJ / mol ( 1.50 mol ) = 61.2 kJ of heat is to be absorbed
Given the constant rate of 19.0 J/s supply of energy to the system, we determine the time as follows:
Time = 61.2 kJ ( 1000 J / 1 kJ ) / 19.0 J/s = 3221.05 s
The proper way to chemically write carbon dioxide is CO2
Answer:
count how many electrons it 'owns'
Answer:
Δ[NH₃]/Δt = 2/3 ( Δ[H₂]/Δt )
Explanation:
For determining rates as a function of reaction coefficients one should realize that these type problems are <u>always in pairs</u> of reaction components. For the reaction N₂ + 3H₂ => 2NH₃ one can compare ...
Δ[N₂]/Δt ∝ Δ[H₂]/Δt, or
Δ[N₂]/Δt ∝ Δ[NH₃]/Δt, or
Δ[H₂]/Δt ∝ Δ[NH₃]/Δt, but never 3 at a time.
So, set up the relationship of interest ( ammonia rate vs. hydrogen rate)... nitrogen rate is ignored.
Δ[H₂]/Δt ∝ Δ[NH₃]/Δt
Now, 'swap' coefficients of balanced equation and apply to terms given then set term multiples equal ...
N₂ + 3H₂ => 2NH₃ => 2(Δ[H₂]/Δt) = 3(Δ[NH₃]/Δt) => 2/3(Δ[H₂]/Δt) = (Δ[NH₃]/Δt)
NOTE => Comparing rates individually of the component rates in reaction process, the rate of H₂(g) consumption is 3/2 times <u>faster</u> than NH₃(g) production (larger coefficient). So, in order to compose an equivalent mathematical relationship between the two, one must reduce the rate of the H₂(g) by 2/3 in order to equal the rate of NH₃(g) production. Now, given the rate of one of the components as a given, substitute and solve for the unknown.
CAUTION => When Interpreting rate of reaction one should note that the rate expression for an individual reaction component defines 'instantaneous' rate or speed. <u>This means velocity (or, speed) does not have 'signage'</u>. Yes, one may say the rate is higher or lower as time changes but that change is an acceleration or deceleration for one instantaneous velocity to another. Acceleration and Deceleration do have signage but the positional instantaneous velocity (defined by a point in time) does not. Thus is reason for the 'e-choice' answer selection without the signage associated with the expression terms.
