Answer:
Mass of original sample = 100 g
Explanation:
Half life of cesium-137 = 30.17 years
Where, k is rate constant
So,
The rate constant, k = 0.02297 year⁻¹
Time = 90.6 years
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Initial concentration = ?
Final concentration = 12.5 grams
Applying in the above equation, we get that:-
![12.5\ g=[A_0]e^{-0.02297\times 90.6}](https://tex.z-dn.net/?f=12.5%5C%20g%3D%5BA_0%5De%5E%7B-0.02297%5Ctimes%2090.6%7D)
![[A_0]=\frac{12.5}{e^{-0.02297\times 90.6}}\ g=100\ g](https://tex.z-dn.net/?f=%5BA_0%5D%3D%5Cfrac%7B12.5%7D%7Be%5E%7B-0.02297%5Ctimes%2090.6%7D%7D%5C%20g%3D100%5C%20g)
<u>Mass of original sample = 100 g</u>
Calcium forms an ion with a positive 2 charge and chlorine forms an ion with a negative one charg, so the formula is <span>CaC<span>l2</span></span>
Group 1 metals and group 2 metals form positive ions by losing 1 and 2 electrons respectively. Non-metals in group 17 gain 1, group 16 gain 2 and group 15 gain 3. Elements which lose electrons form positive ions while elements that gain electrons form negative ions.
To write a formula, you must balance charges so the overall charge is zero. A simple way to do this is to swap the # of the ion's charge and make it the subscript of the other ion. However, leave off the number 1 and reduce to lowest whole number ratio.
<span>This was one of the old regents. I didn't get this at first glance too so I searched for it and I got this site. lol. I know what to do now. To get a higher boiling point, you have to find which solution has the most moles. Since SO4, PO4, and NO3 are all polyatomic ions, they only count as one mole. </span>
Answer:
0.499atm
Explanation:
The formula is
P1/V1 = P2/V2
so:
1.26atm/7.40L = P2/2.93L
then:
(1.26atm/7.40L)*2.93L = P2
= 0.4988918911atm
the answer must have 3 sig figs
