I think it can be dissolved but if u keep adding more sugar it will stop dissolving
Answer:
Check the explanation
Explanation:
AT = A0 e(-T/H)
... where A0 is the starting activity, AT is the activity at some time T, and H is the half-life, in units of T.
Substituting what we know, we get...
0.71 = (1) e(-T/5730)
Solve for T...
loge(0.71) = -T/5730
T = -loge(0.71)(5730)
T = 1962 (conservatively rounded, T = 2000)
similarly for all
for aboriginal charcoal
0.28 = (1) e(-T/5730)
Solve for T...
loge(0.28) = -T/5730
T = -loge(0.28)(5730)
T = 7294 (conservatively rounded, T = 7000)
for mayan headdress
0.89 = (1) e(-T/5730)
Solve for T...
loge(0.89) = -T/5730
T = -loge(0.89)(5730)
T = 667 (conservatively rounded, T = 700)
for neanderthal
0.05 = (1) e(-T/5730)
Solve for T...
loge(0.05) = -T/5730
T = -loge(0.05)(5730)
T = 17165 (conservatively rounded, T = 17000)
Answer:
The initial rate of the reaction between substances P and Q was measured in a series of
experiments and the following rate equation was deduced.
![rate = k[P]^{2} [Q]](https://tex.z-dn.net/?f=rate%20%3D%20k%5BP%5D%5E%7B2%7D%20%5BQ%5D)
Complete the table of data below for the reaction between P and Q
Explanation:
Given rate of the reaction is:
![rate= k[P]^{2} [Q]\\=>[Q]=\frac{rate}{k.[P]^{2} } \\and \\\\\\\ [P]=\sqrt{\frac{rate}{k.[Q]} }](https://tex.z-dn.net/?f=rate%3D%20k%5BP%5D%5E%7B2%7D%20%5BQ%5D%5C%5C%3D%3E%5BQ%5D%3D%5Cfrac%7Brate%7D%7Bk.%5BP%5D%5E%7B2%7D%20%7D%20%5C%5Cand%20%5C%5C%5C%5C%5C%5C%5C%20%5BP%5D%3D%5Csqrt%7B%5Cfrac%7Brate%7D%7Bk.%5BQ%5D%7D%20%7D)
Substitute the given values in this formulae to get the [P], [Q] and rate values.
From the first row,
the value of k can be calulated:
![k=\frac{rate}{[P]^{2}[Q] } \\ =\frac{4.8*10^-3}{(0.2)^{2} 2. (0.30)} \\ =0.4](https://tex.z-dn.net/?f=k%3D%5Cfrac%7Brate%7D%7B%5BP%5D%5E%7B2%7D%5BQ%5D%20%7D%20%5C%5C%20%20%3D%5Cfrac%7B4.8%2A10%5E-3%7D%7B%280.2%29%5E%7B2%7D%202.%20%280.30%29%7D%20%5C%5C%20%3D0.4)
Second row:
2. Rate value:
3.Third row:
![[Q]=\frac{rate}{k.[P]^{2} } \\ =9.6*10^-3 / (0.4 *(0.40)^{2} \\ =0.15mol.dm^{-3}](https://tex.z-dn.net/?f=%5BQ%5D%3D%5Cfrac%7Brate%7D%7Bk.%5BP%5D%5E%7B2%7D%20%7D%20%5C%5C%20%20%20%20%20%3D9.6%2A10%5E-3%20%2F%20%280.4%20%2A%280.40%29%5E%7B2%7D%20%5C%5C%20%20%20%20%3D0.15mol.dm%5E%7B-3%7D)
4. Fourth row:
![[P]=\sqrt{\frac{rate}{k.[Q]} }\\=>[P]=\sqrt{\frac{19.2*10^-3}{0.60*0.4} } \\=>[P]=0.283mol.dm^{-3}](https://tex.z-dn.net/?f=%5BP%5D%3D%5Csqrt%7B%5Cfrac%7Brate%7D%7Bk.%5BQ%5D%7D%20%7D%5C%5C%3D%3E%5BP%5D%3D%5Csqrt%7B%5Cfrac%7B19.2%2A10%5E-3%7D%7B0.60%2A0.4%7D%20%7D%20%5C%5C%3D%3E%5BP%5D%3D0.283mol.dm%5E%7B-3%7D)
Answer:
<em>Zinc nitrate is an inorganic chemical compound with the formula Zn(NO3)2. This white, crystalline salt is highly deliquescent and is typically encountered as a hexahydrate Zn(NO3)2•6H2O. It is soluble in both water and alcohol.</em>
Explanation:
correct me if im wrong please
Your answer is 3.25 moles of Bromine
