Answer:
6.12 L
Explanation:
Given that,
Initial volume, V₁ = 5 L
Initial temperature, T₁ = 7.0°C = 343 K
Final temperature, T₂ = 147°C = 420 K
We need to find its new volume. The relation between volume and temperature is given by :
So, the new volume is 6.12 L.
Answer:
a. 300 Kg of fertilizer
b. 225 Kg of fertilizer
c. 400 Kg of fertilizer
d. 600 Kg of fertilizer
Explanation:
The percentage composition ratio of Nitrogen, Phosphorus and Potassium in a 1 Kg bag of the given fertilizer is 40:15:10.
Therefore a 1 Kg bag contains;
40/100 * 1 Kg = 0.4 Kg of Nitrogen;
15/100 * 1 Kg = 0.15 Kg of phosphorus;
10/100 * 1 Kg = 0.1 Kg of potassium
Quantity of fertilizer required to add to a hectare to supply;
a. Nitrogen at 120 kg/ha = 120/0.4 = 300 Kg of fertilizer
b.. Nitrogen at 90 Kg/ha = 90/0.4 = 225 Kg of fertilizer
c. Phosphorus at a rate of 60 kg/ha = 60/0.15 = 400 Kg of fertilizer
d. Potassium at a rate of 60 kg/ha = 60/0.1 = 600 Kg of fertilizer
Answer:
a) First-order.
b) 0.013 min⁻¹
c) 53.3 min.
d) 0.0142M
Explanation:
Hello,
In this case, on the attached document, we can notice the corresponding plot for each possible order of reaction. Thus, we should remember that in zeroth-order we plot the concentration of the reactant (SO2Cl2 ) versus the time, in first-order the natural logarithm of the concentration of the reactant (SO2Cl2 ) versus the time and in second-order reactions the inverse of the concentration of the reactant (SO2Cl2 ) versus the time.
a) In such a way, we realize the best fit is exhibited by the first-order model which shows a straight line (R=1) which has a slope of -0.0013 and an intercept of -2.3025 (natural logarithm of 0.1 which corresponds to the initial concentration). Therefore, the reaction has a first-order kinetics.
b) Since the slope is -0.0013 (take two random values), the rate constant is 0.013 min⁻¹:
c) Half life for first-order kinetics is computed by:
d) Here, we compute the concentration via the integrated rate law once 1500 minutes have passed:
Best regards.
Answer:
Q = 3139.5 j
Explanation:
Given data:
Mass = 50 g
Initial temperature = 25°C
Final temperature = 95°C
Specific heat capacity = 0.897 j/g.°C
Heat absorbed = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = T2 - T1
ΔT = 95°C - 25°C
ΔT = 70°C
Q = m.c. ΔT
Q = 50 g× 0.897 J/g.°C ×70°C
Q = 3139.5 j
Answer:
The solution warmed and the solubility of Q increased
Explanation: