Answer:
the answer would be an ever-growing amount of red lion-fish. and could result in some extinct species in a pristine reef.
Explanation: all in all people would get rid of the fish before the whole place gets rect.
Answer:
3626.76dm³
Explanation:
Given parameters:
Number of moles of Nitrogen in tank = 17moles
Temperature of the gas = 34°C
Pressure on the gas = 12000Pa
Unkown:
Volume of the tank, V =?
Converting the parameters to workable units:
We take the temperature from °C to Kelvin
K = 273 + °C = 273 + 34 = 307k
Taking the pressure in Pa to atm:
101325Pa = 1atm
12000Pa = 0.118atm
Solution:
To solve this problem, we employ the use of the ideal gas equation. The ideal gas law combines three gas laws which are the Boyle's law, Charles's law and the Avogadro's law.
It is expressed as PV = nRT
The unknown is the Volume and we make it the subject of the formula
V = 
Where R is called the gas constant and it is given as 0.082atmdm³mol⁻¹K⁻¹
Therefore V =
= 3626.76dm³
Answer: Honey has a much lower vapor pressure than pure water has. So, pure water evaporates at a much higher rate.
Explanation:
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.
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)