This is an ideal gas law question. It uses the ideal gas law equation, PV =nRT.
P = pressure, V = volume, n = moles, R is the constant, and T is temperature in kelvin. The temperature needs to be converted Kevin first. To convert from Celsius to Kevin, you add 273, meaning that the temp in Kevin is 403K. Then plug all the info into the equation to solve for moles.
(1.00atm)(1.280L)=n(0.0821)(403K)
n = 0.0387moles
To find molar mass, divide mass by moles.
4.03g / 0.0387moles = 104.17g/mol
104g/mol rounded to three significant digits
Its C. It has to be equivalent to the number divided or multiplied by 5.
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)
Answer:
v = 2,66x10⁻⁵ P[H₂C₂O₄]
Explanation:
For the reaction:
H₂C₂O₄(g) → CO₂(g) + HCOOH(g)
At t = 0, the initial pressure is just of H₂C₂O₄(g). At t= 20000 s, pressures will be:
H₂C₂O₄(g) = P₀ - x
CO₂(g) = x
HCOOH(g) = x
P at t=20000 is:
P₀ - x + x + x = P₀+x. That means P at t=20000s - P₀ = x
For 1st point:
x = 92,8-65,8 = 27
Pressure of H₂C₂O₄(g) at t=20000s: 65,8-27 = 38,8
2nd point:
x = 130-92,1 = 37,9
H₂C₂O₄(g): 92,1 - 37,9 = 54,2
3rd point:
x = 157-111 = 46
H₂C₂O₄(g): 111-46 = 65
Now, as the rate law is :
v = k P[H₂C₂O₄]
Based on integrated rate law, k is:
(- ln P[H₂C₂O₄] + ln P[H₂C₂O₄]₀) / t = k
1st point:
k = 2,64x10⁻⁵
2nd point:
k = 2,65x10⁻⁵
3rd point:
k = 2,68x10⁻⁵
The averrage of this values is:
k = 2,66x10⁻⁵
That means law is:
v = 2,66x10⁻⁵ P[H₂C₂O₄]
I hope it helps!
Sugar, on the other hand, is composed of carbon, oxygen, and hydrogen and has covalent bonds. A salt molecule is made up of one sodium atom and one chlorine atom.