Answer:
Explanation:
We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.
The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.
- ΔT= final temperature - initial temperature
The sample was heated <em>from </em> 58.8 degrees Celsius to 88.9 degrees Celsius.
- ΔT= 88.9 °C - 58.8 °C = 30.1 °C
Now we know three variables:
- Q= 4500.0 J
- c= 0.4494 J/g°C
- ΔT = 30.1 °C
Substitute these values into the formula.
Multiply on the right side of the equation. The units of degrees Celsius cancel.
We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g
The units of Joules cancel.
The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.
The mass of the sample of metal is approximately <u>333 grams.</u>
Answer
7665 years
Procedure
Let N₀ be the amount of carbon-14 present in a living organism. According to the radioactive decay law, the number of carbon-14 atoms, N, left in a dead tissue sample after a certain time, t, is given by the exponential equation:
N = N₀e^(-λt)
where λ is the decay constant which is related to half-life (T1/2) by the equation:
Here, ln(2) is the natural logarithm of 2.
The percent of carbon-14 remaining after time t is given by N/N₀.
Using the first equation, we can determine λt.
The half-life of carbon-14 is 5,720 years, thus, we can calculate λ using the second equation, and then find t.
Solving the second equation for t, and using the λ we have just calculated we will have
t= 7665 years
Answer:Answer: (B) Archaebacteria
Explanation:
Explanation: Archaebacteria, like all prokaryotes, have no membrane bound organelles. This means that the archaebacteria are without nuclei, mitochondria, endoplasmic reticula, lysosomes, Golgi complexes, or chloroplasts. Because these organisms have no nucleus, the genetic material floats freely in the cytoplasm
Answer:
b. Second order in NO and first order in O₂.
Explanation:
A. The mechanism
![\rm 2NO\xrightarrow[k_{-1}]{k_{1}}N_{2}O_{2} \, (fast)\\\rm N_{2}O_{2} + O_{2}\xrightarrow{k_{2}} 2NO_{2} \, (slow)](https://tex.z-dn.net/?f=%5Crm%202NO%5Cxrightarrow%5Bk_%7B-1%7D%5D%7Bk_%7B1%7D%7DN_%7B2%7DO_%7B2%7D%20%5C%2C%20%28fast%29%5C%5C%5Crm%20N_%7B2%7DO_%7B2%7D%20%2B%20O_%7B2%7D%5Cxrightarrow%7Bk_%7B2%7D%7D%202NO_%7B2%7D%20%5C%2C%20%28slow%29)
B. The rate expressions
![-\dfrac{\text{d[NO]} }{\text{d}t} = k_{1}[\text{NO]}^{2} - k_{-1} [\text{N}_{2}\text{O}_{2}]^{2}\\\\\rm -\dfrac{\text{d[N$_{2}$O$_{2}$]}}{\text{d}t} = -\dfrac{\text{d[O$_{2}$]}}{\text{d}t} = k_{2}[ N_{2}O_{2}][O_{2}] - k_{1} [NO]^{2}\\\\\dfrac{\text{d[NO$_{2}$]}}{\text{d}t}= k_{2}[ N_{2}O_{2}][O_{2}]](https://tex.z-dn.net/?f=-%5Cdfrac%7B%5Ctext%7Bd%5BNO%5D%7D%20%7D%7B%5Ctext%7Bd%7Dt%7D%20%3D%20k_%7B1%7D%5B%5Ctext%7BNO%5D%7D%5E%7B2%7D%20-%20k_%7B-1%7D%20%5B%5Ctext%7BN%7D_%7B2%7D%5Ctext%7BO%7D_%7B2%7D%5D%5E%7B2%7D%5C%5C%5C%5C%5Crm%20-%5Cdfrac%7B%5Ctext%7Bd%5BN%24_%7B2%7D%24O%24_%7B2%7D%24%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%20%3D%20-%5Cdfrac%7B%5Ctext%7Bd%5BO%24_%7B2%7D%24%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%20%3D%20k_%7B2%7D%5B%20N_%7B2%7DO_%7B2%7D%5D%5BO_%7B2%7D%5D%20-%20k_%7B1%7D%20%5BNO%5D%5E%7B2%7D%5C%5C%5C%5C%5Cdfrac%7B%5Ctext%7Bd%5BNO%24_%7B2%7D%24%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%3D%20k_%7B2%7D%5B%20N_%7B2%7DO_%7B2%7D%5D%5BO_%7B2%7D%5D)
The last expression is the rate law for the slow step. However, it contains the intermediate N₂O₂, so it can't be the final answer.
C. Assume the first step is an equilibrium
If the first step is an equilibrium, the rates of the forward and reverse reactions are equal. The equilibrium is only slightly perturbed by the slow leaking away of N₂O₂ to form product.
![\rm k_{1}[NO]^{2} = k_{-1} [N_{2}O_{2}]\\\\\rm [N_{2}O_{2}] = \dfrac{k_{1}}{k_{-1}}[NO]^{2}](https://tex.z-dn.net/?f=%5Crm%20k_%7B1%7D%5BNO%5D%5E%7B2%7D%20%3D%20k_%7B-1%7D%20%5BN_%7B2%7DO_%7B2%7D%5D%5C%5C%5C%5C%5Crm%20%5BN_%7B2%7DO_%7B2%7D%5D%20%3D%20%5Cdfrac%7Bk_%7B1%7D%7D%7Bk_%7B-1%7D%7D%5BNO%5D%5E%7B2%7D)
D. Substitute this concentration into the rate law
![\rm \dfrac{\text{d[NO$_{2}$]}}{\text{d}t}= \dfrac{k_{2}k_{1}}{k_{-1}}[NO]^{2} [O_{2}] = k[NO]^{2} [O_{2}]](https://tex.z-dn.net/?f=%5Crm%20%5Cdfrac%7B%5Ctext%7Bd%5BNO%24_%7B2%7D%24%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%3D%20%5Cdfrac%7Bk_%7B2%7Dk_%7B1%7D%7D%7Bk_%7B-1%7D%7D%5BNO%5D%5E%7B2%7D%20%5BO_%7B2%7D%5D%20%3D%20k%5BNO%5D%5E%7B2%7D%20%5BO_%7B2%7D%5D)
The reaction is second order in NO and first order in O₂.
Answer:
Chlorine
Explanation:
Even though chlorine is highly electronegative, the best answer is no, and in this class we will consider chlorine not to form hydrogen bonds (even though it has the same electronegativity as oxygen). This is because chlorine is large and its lone electron is in a diffuse orbital, covering a large area, and thus do not have the high charge density to act as a strong hydrogen bond acceptor. But it does form weak hydrogen bonds in solid crystalline hydrogen chloride at very low temperatures.
