<span>The half-life of Carbon 14 and radionuclides are used to estimate the absolute (versus relative) age of pre-history items </span>
Answer:
volume of the gas is 5.0L
Explanation:
Using Boyle's law that state the pressure of a gas is inversely proportional to volume of it occupies when temperature is constant, it is possible to write:
P₁V₁ = P₂V₂
<em>Where P is pressure, V is volume and 1 and 2 are initial and final states.</em>
<em />
If initial volume is 2.5L, initial pressure is 2.0atm and 1.0atm is final pressure, final volume is:
2.0atm*2.5L = 1atm V₂
5.0L = V₂
Thus, <em>volume of the gas is 5.0L</em>.
Answer:
Oxidation number:
3*1+ oxidation number of J+2*-2= -1
Oxidation number of J = 0
Answer:
Half-life = 3 minutes
Explanation:
Using the radioactive decay equation we can solve for reaction constant, k. And by using:
K = ln2 / Half-life
We can find half-life of polonium-218
Radioactive decay:
Ln[A] = -kt + ln [A]₀
Where:
[A] could be taken as mass of polonium after t time: 1.0mg
k is Reaction constant, our incognite
t are 12 min
[A]₀ initial amount of polonium-218: 16mg
Ln[A] = -kt + ln [A]₀
Ln[1.0mg] = -k*12min + ln [16mg]
-2.7726 = - k*12min
k = 0.231min⁻¹
Half-life = ln 2 / 0.231min⁻¹
<h3>Half-life = 3 minutes</h3>
Answer:
1. 2.510kJ
2. Q = 1.5 kJ
Explanation:
Hello there!
In this case, according to the given information for this calorimetry problem, we can proceed as follows:
1. Here, we consider the following equivalence statement for converting from calories to joules and from joules to kilojoules:
![1cal=4.184J\\\\1kJ=1000J](https://tex.z-dn.net/?f=1cal%3D4.184J%5C%5C%5C%5C1kJ%3D1000J)
Then, we perform the conversion as follows:
![600.0cal*\frac{4.184J}{1cal}*\frac{1kJ}{1000J}=2.510kJ](https://tex.z-dn.net/?f=600.0cal%2A%5Cfrac%7B4.184J%7D%7B1cal%7D%2A%5Cfrac%7B1kJ%7D%7B1000J%7D%3D2.510kJ)
2. Here, we use the general heat equation:
![Q=mC(T_2-T_1)](https://tex.z-dn.net/?f=Q%3DmC%28T_2-T_1%29)
And we plug in the given mass, specific heat and initial and final temperature to obtain:
![Q=236g*0.24\frac{J}{g\°C} (34.9\°C-8.5\°C)\\\\Q=1495.3J*\frac{1kJ}{1000J} \\\\Q=1.5kJ](https://tex.z-dn.net/?f=Q%3D236g%2A0.24%5Cfrac%7BJ%7D%7Bg%5C%C2%B0C%7D%20%2834.9%5C%C2%B0C-8.5%5C%C2%B0C%29%5C%5C%5C%5CQ%3D1495.3J%2A%5Cfrac%7B1kJ%7D%7B1000J%7D%20%5C%5C%5C%5CQ%3D1.5kJ)
Regards!