Answer:
D. 40 % increase
Step-by-step explanation:
r = k[A]²/[B]
The rate is inversely proportional to [B]. If [B] is doubled, the rate is halved.
We must double this rate to get back to the original.
The rate is directly proportional to [A]².
2 = [A]₂/[A]₁² Take the square root of each side
√2 = [A]₂/[A]₁ Multiply each side by [A]₁
[A]₂ = √2[A]₁
[A]₂ = 1.41[A]₁
We must increase [A] by 41 %.
Answer:
349.22°C
Explanation:
Let the final temperature of the two pieces of metal be x.
Now, the warmer metal which is C u reduces from 475°C to x. Thus Δt for C u is; Δt1 = 475 - x.
The cooler metal Cr increases in temperature from 265°C to x. Thus, it's change in temperature is Δt for Cr is; Δt2 = x - 265.
Now from conservation of energy, the amount of energy leaving the C u metal is equal to the amount of energy entering the Cr metal.
Thus;
q_lost = q_gain
Where;
q_lost = m1•c1•Δt1
q_gained = m2•c2•Δt2
Now, c1 & c2 are the specific heat capacity of C u and Cr respectively.
From online tables, c1 = 0.385 J/g°C and c2 = 0.46 J/g°C
We are given;
m1 = 12g and m2 = 15g
Thus;
12 × 0.385 × (475 - x) = 15 × 0.46 × (x - 265)
2194.5 - 4.62x = 6.9x - 1828.5
6.9x + 4.62x = 2194.5 + 1828.5
11.52x = 4023
x = 4023/11.52
x = 349.22°C
Answer:
Initial pressure = 157 kpa (Approx)
Explanation:
Given:
final temperature = 234 K
final pressure = 210 kpa
Initial temperature = 175 K
Find:
Initial pressure
Computation:
Initial pressure / Initial temperature = final pressure / final temperature
Initial pressure / 175 = 210 / 234
Initial pressure = 157 kpa (Approx)
Answer : The energy removed must be, -67.7 kJ
Solution :
The process involved in this problem are :
The expression used will be:
![\Delta H=[m\times c_{p,g}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,l}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bm%5Ctimes%20c_%7Bp%2Cg%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bm%5Ctimes%20%5CDelta%20H_%7Bvap%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= heat released by the reaction = ?
m = mass of benzene = 125 g
= specific heat of gaseous benzene = 
= specific heat of liquid benzene = 
= enthalpy change for vaporization = 
Molar mass of benzene = 78.11 g/mole
Now put all the given values in the above expression, we get:
![\Delta H=[125g\times 1.06J/g.K\times (353.0-(425.0))K]+125g\times -434.0J/g+[125g\times 1.73J/g.K\times (335.0-353.0)K]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B125g%5Ctimes%201.06J%2Fg.K%5Ctimes%20%28353.0-%28425.0%29%29K%5D%2B125g%5Ctimes%20-434.0J%2Fg%2B%5B125g%5Ctimes%201.73J%2Fg.K%5Ctimes%20%28335.0-353.0%29K%5D)
Therefore, the energy removed must be, -67.7 kJ
Answer:
Acting how bees act. I had that question. :3.
