The cell theory states that all cells come from other cells. When cells reproduce, they make copies of their DNA and make copies of their DNA and pass it on to the new cells.
Answer:
55
Explanation:
25g is less dense than 80g therefore will mostly float. if you subtract 80-25 that will leave you will 55 as the difference
Answer:
![\%LiClO_3=90.4\%](https://tex.z-dn.net/?f=%5C%25LiClO_3%3D90.4%5C%25)
Explanation:
Hello!
In this case, since the undergoing chemical reaction is:
![LiClO_4\rightarrow LiCl+\frac{3}{2} O_2](https://tex.z-dn.net/?f=LiClO_4%5Crightarrow%20LiCl%2B%5Cfrac%7B3%7D%7B2%7D%20O_2)
It is widely known that when a gas given off from a reaction is collected over water, we can compute its pressure by minusing the total pressure 762 mmHg and the vapor pressure of water at the experiment's temperature (20 °C) in this case 17.5 mmHg as shown below:
![p_{O_2}=762mmHg-17.5mmHg=744.5mmHg*\frac{1atm}{760mmHg}=0.980atm](https://tex.z-dn.net/?f=p_%7BO_2%7D%3D762mmHg-17.5mmHg%3D744.5mmHg%2A%5Cfrac%7B1atm%7D%7B760mmHg%7D%3D0.980atm)
Next, by using the ideal gas equation we compute the yielded moles of oxygen considering the collected 313 mL (0.313 L):
![n_{O_2}=\frac{PV}{RT}=\frac{0.980atm*0.313L}{0.082\frac{atm*L}{mol*K}*293.15K}\\\\ n_{O_2}=0.0128molO_2](https://tex.z-dn.net/?f=n_%7BO_2%7D%3D%5Cfrac%7BPV%7D%7BRT%7D%3D%5Cfrac%7B0.980atm%2A0.313L%7D%7B0.082%5Cfrac%7Batm%2AL%7D%7Bmol%2AK%7D%2A293.15K%7D%5C%5C%5C%5C%20%20n_%7BO_2%7D%3D0.0128molO_2)
Now, via the 3/2:1 mole ratio between oxygen and lithium chlorate (molar mass = 90.39 g/mol), we compute the original mass of decomposed lithium chlorate as follows:
![m_{LiClO_3}=0.0128molO_2*\frac{1molLiClO_3}{3/2molO_2}*\frac{90.39gLiClO_3}{1molLiClO_3} \\\\m_{LiClO_3}=0.771gLiClO_3](https://tex.z-dn.net/?f=m_%7BLiClO_3%7D%3D0.0128molO_2%2A%5Cfrac%7B1molLiClO_3%7D%7B3%2F2molO_2%7D%2A%5Cfrac%7B90.39gLiClO_3%7D%7B1molLiClO_3%7D%20%20%5C%5C%5C%5Cm_%7BLiClO_3%7D%3D0.771gLiClO_3)
Now, the percentage is computed as shown below:
![\%LiClO_3=\frac{0.771g}{0.853g} *100\%\\\\\%LiClO_3=90.4\%](https://tex.z-dn.net/?f=%5C%25LiClO_3%3D%5Cfrac%7B0.771g%7D%7B0.853g%7D%20%2A100%5C%25%5C%5C%5C%5C%5C%25LiClO_3%3D90.4%5C%25)
Best regards!
Answer:
Explanation:
2 HgCl₂ + C₂O₄²⁻ = 2 Cl⁻ + Hg₂Cl₂ + 2CO₂
1 )
Rate of reaction ![= k[HgCl_2]^m[C_2O_4^{-2}]^n](https://tex.z-dn.net/?f=%3D%20k%5BHgCl_2%5D%5Em%5BC_2O_4%5E%7B-2%7D%5D%5En)
[HgCl₂] [C₂O₄²⁻ ] Rate
1 . .124 .115 1.61 x 10⁻⁵
2 . .248 .115 3.23 x 10⁻⁵
3 . .124 .229 6.4 x 10⁻⁵
4 . .248 .229 1.28 x 10⁻⁴
comparing 1 and 3 , when concentration of HgCl₂ remains constant and concentration of C₂O₄²⁻ becomes twice , rate becomes 4 times so rate is proportional to square of concentration of C₂O₄²⁻ .
Hence n = 2
comparing 1 and 2 , when concentration of HgCl₂ becomes twice and concentration of C₂O₄²⁻ remains constant , rate becomes 2 times so rate is proportional to simply concentration of C₂O₄²⁻ .
Hence m = 1
Putting the data of 1 in the rate equation found
1.61 x 10⁻⁵ = k x .124 x .115²
k = 11.3 x 10⁻⁴ M⁻² s⁻¹
Answer:
I guess number three (3) Is wrong