The nuclides 16O8 and 15O8 differs in their atomic weight. 16O8 has an atomic weight of 16 amu while 15O8 has an atomic weight of 15 amu. They also differ in the amount of neutrons, 16O8 has 8 protons and 8 neutrons (16 - 8) while 15O8 only has 7 neutrons (15 - 8 protons). <span />
Answer:
![[SO_2Cl_2]_{600}= 0.0842 M](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D_%7B600%7D%3D%200.0842%20M)
Explanation:
Some theoretical knowledge is required here. We should understand that whenever we plot the natural logarithm, ln, of a concentration vs. time and obtain a straight line, this indicates a first-order reaction. That said, since this is the case here, we have a first-order reaction with respect to
.
The linear equation has the following terms:

It is a linear form of the integrated first-order law equation:
![ln[SO_2Cl_2]_t = -kt + ln[SO_2Cl_2]_o](https://tex.z-dn.net/?f=ln%5BSO_2Cl_2%5D_t%20%3D%20-kt%20%2B%20ln%5BSO_2Cl_2%5D_o)
Therefore, the rate constant, k, is:

The natural logarithm of initial molarity is:
![ln[SO_2Cl_2]_o = -2.30](https://tex.z-dn.net/?f=ln%5BSO_2Cl_2%5D_o%20%3D%20-2.30)
Using the equation, we may substitute for t = 600 s and obtain the natural logarithm of the concentration at that time:
![ln[SO_2Cl_2]_{600} = -0.000290 s^{-1}\cdot 600 s - 2.30 = -2.474](https://tex.z-dn.net/?f=ln%5BSO_2Cl_2%5D_%7B600%7D%20%3D%20-0.000290%20s%5E%7B-1%7D%5Ccdot%20600%20s%20-%202.30%20%3D%20-2.474)
Take the antilog of both sides to find the actual molarity:
![[SO_2Cl_2]_{600}=e^{-2.474} = 0.0842 M](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D_%7B600%7D%3De%5E%7B-2.474%7D%20%3D%200.0842%20M)
Possibly different But here I hope this helps Recall that neutrons, along with protons, make up the nucleus. Since neutrons have no electrical charge, they do not affect the chemical behavior. ... Atoms of the same element with different numbers of neutrons are called isotopes.
Answer:
9 × 10⁻³ mol·L⁻¹s⁻¹
Explanation:
Data:
k = 1 × 10⁻³ L·mol⁻¹s⁻¹
[A] = 3 mol·L⁻¹
Calculation:
rate = k[A]² = 1 × 10⁻³ L·mol⁻¹s⁻¹ × (3 mol·L⁻¹)² = 9 × 10⁻³ mol·L⁻¹s⁻¹
4.8 g/cm3 with sig figs since it's mass/volume you divide 76 grams by 16 cm3