Extensive properties depend on the amount of matter present, for example, the mass of gold. Intensive properties do not depend on the amount of matter present, for example, the density of gold. Heat is an example of an extensive property, and temperature is an example of an intensive property.
I would say
D. Gravity pulled particles of dust and gas together to form planets.Hope it helped!
Compounds are not formed by accepting electrons. Number 1 and 3 are correct.
Answer:
B) 1270 torr
Explanation:
Given data
- Initial volume (V₁): 5.00 L
- Initial pressure (P₁): 760 torr
- Final volume (V₂): 3.00 L
We can find the final pressure using Boyle's law.
P₁ × V₁ = P₂ × V₂
P₂ = P₁ × V₁/V₂
P₂ = 760 torr × 5.00 L/3.00 L
P₂ = 1.27 × 10³ torr = 1270 torr
The final pressure is 1270 torr.
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.