Most atoms have three different subatomic particles inside them: protons, neutrons, and electrons. The protons and neutrons are packed together into the center of the atom(which is called the nucleus) and the electrons, which are very much smaller, whizz around the outside. Most of an atom is empty space.
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Answer:
Mole fraction O₂= 0.43
Explanation:
Mole fraction is the moles of gas/ total moles.
Let's determine the moles of each:
Moles O₂ → 15.1 g / 16 g/mol = 0.94
Moles N₂ → 8.19 g / 14 g/mol = 0.013
Moles H₂ → 2.46 / 2 g/mol = 1.23
Total moles = 2.183
Mole fraction O₂= 0.94 / 2.183 → 0.43
By adding together the number of protons and neutrons and multiplying by 1 amu, you can calculate the mass of the atom.
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)
Answer:
b. unsaturated
.
Explanation:
Hello there!
In this case, according to the given information, it turns out necessary for us to bear to mind the definition of each type of solution:
- Supersaturated solution: comprises a large amount of solute at a temperature at which it will be able to crystalize upon standing.
- Unsaturated solution: is a solution in which a solvent is able to dissolve any more solute at a given temperature.
- Saturated solution can be defined as a solution in which a solvent is not capable of dissolving any more solute at a given temperature.
In such a way, since 20 grams of the solute are less than the solubility, we infer this is b. unsaturated, as 33.3 grams of solute can be further added to the 100 grams of water.
Regards!