Answer:
The atomic mass of the boron atom would be <em>10.135</em>
Explanation:
This is generally known as relative atomic mass.
Relative atomic mass or atomic weight is a physical quantity defined as the ratio of the average mass of atoms of a chemical element in a given sample to the atomic mass of 1/12 of the mass of a carbon-12 atom. Since both quantities in the ratio are masses, the resulting value is dimensionless; hence the value is said to be relative and does not have a unit.
<em>Note that the relative atomic mass of atoms is not always a whole number because of it being isotopic in nature.</em>
- <em>Divide each abundance by 100 then multiply by atomic mass</em>
- <em>Do that for each isotope, then add the two result. Thus</em>
Relative atomic mass of Boron = (18.5/100 x 11) + (81/100 x 10)
= 2.035 + 8.1
= 10.135
Answer:
(a) 
(b) 
(c) 
(d) 
Explanation:
Hello,
In this case, given the solubility of each salt, we can compute their molar solubilities by using the molar masses. Afterwards, by using the mole ratio between ions, we can compute the concentration of each dissolved and therefore the solubility product:
(a) 
In such a way, as barium and selenate ions are in 1:1 molar ratio, they have the same concentration, for which the solubility product turns out:
![Ksp=[Ba^{2+}][SeO_4^{2-}]=(6.7x10^{-4}\frac{mol}{L} )^2\\\\Ksp=4.50x10^{-7}](https://tex.z-dn.net/?f=Ksp%3D%5BBa%5E%7B2%2B%7D%5D%5BSeO_4%5E%7B2-%7D%5D%3D%286.7x10%5E%7B-4%7D%5Cfrac%7Bmol%7D%7BL%7D%20%20%20%29%5E2%5C%5C%5C%5CKsp%3D4.50x10%5E%7B-7%7D)
(B) 
In such a way, as barium and bromate ions are in 1:2 molar ratio, bromate ions have twice the concentration of barium ions, for which the solubility product turns out:
![Ksp=[Ba^{2+}][BrO_3^-]^2=(7.30x10^{-3}\frac{mol}{L})(3.65x10^{-3}\frac{mol}{L})^2\\\\Ksp=1.55x10^{-6}](https://tex.z-dn.net/?f=Ksp%3D%5BBa%5E%7B2%2B%7D%5D%5BBrO_3%5E-%5D%5E2%3D%287.30x10%5E%7B-3%7D%5Cfrac%7Bmol%7D%7BL%7D%29%283.65x10%5E%7B-3%7D%5Cfrac%7Bmol%7D%7BL%7D%29%5E2%5C%5C%5C%5CKsp%3D1.55x10%5E%7B-6%7D)
(C) 
In such a way, as ammonium, magnesium and arsenate ions are in 1:1:1 molar ratio, they have the same concentrations, for which the solubility product turns out:
![Ksp=[NH_4^+][Mg^{2+}][AsO_4^{3-}]^2=(1.31x10^{-4}\frac{mol}{L})^3\\\\Ksp=2.27x10^{-12}](https://tex.z-dn.net/?f=Ksp%3D%5BNH_4%5E%2B%5D%5BMg%5E%7B2%2B%7D%5D%5BAsO_4%5E%7B3-%7D%5D%5E2%3D%281.31x10%5E%7B-4%7D%5Cfrac%7Bmol%7D%7BL%7D%29%5E3%5C%5C%5C%5CKsp%3D2.27x10%5E%7B-12%7D)
(D) 
In such a way, as the involved ions are in 2:3 molar ratio, La ion is twice the molar solubility and MoOs ion is three times it, for which the solubility product turns out:
![Ksp=[La^{3+}]^2[MoOs^{-2}]^3=(2*1.58x10^{-5}\frac{mol}{L})^2(3*1.58x10^{-5}\frac{mol}{L})^3\\\\Ksp=1.05x10^{-22}](https://tex.z-dn.net/?f=Ksp%3D%5BLa%5E%7B3%2B%7D%5D%5E2%5BMoOs%5E%7B-2%7D%5D%5E3%3D%282%2A1.58x10%5E%7B-5%7D%5Cfrac%7Bmol%7D%7BL%7D%29%5E2%283%2A1.58x10%5E%7B-5%7D%5Cfrac%7Bmol%7D%7BL%7D%29%5E3%5C%5C%5C%5CKsp%3D1.05x10%5E%7B-22%7D)
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1N to the right s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s ss s s s
<h3>
Answer:</h3>
3.6 × 10^24 molecules
<h3>
Explanation:</h3>
Concept: Moles and Avogadro's number
- We need to know that one mole of a compound has molecules equivalent to the Avogadro's number, 6.022 × 10^23.
- Therefore; 1 mole of a compound contains 6.022 × 10^23 molecules
In this case;
We are given 6 moles of Methane;
But 1 mole of methane = 6.022 × 10^23 molecules
Therefore;
Molecules = Number of moles × Avogadro's number
= 6 moles × 6.022 × 10^23 molecules/mole
= 3.613 × 10^24 molecules
= 3.6× 10^24 molecules
Thus, there are 3.6× 10^24 molecules in 6 moles of methane.
