Answer:
You can fill 212 balloons.
Explanation:
First we <u>calculate the helium moles in the small cylinder</u>, using <em>PV=nRT:</em>
- P = 14300 kPa ⇒ 14300 * 0.009869 = 141.13 atm
- R = 0.082 atm·L·mol⁻¹·K⁻¹
- T = 25 °C ⇒ 25 + 273.16 = 298.16 K
141.13 atm * 2.20 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K
Then we <u>calculate the number of moles that can fit in a single balloon</u>:
- 1.22 atm * 1.20 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K
Finally we <u>divide the total number of available moles by the number of moles in a single balloon</u>:
- 12.70 mol / 0.0599 mol = 212.09
So the answer is that you can fill 212 balloons.
<span>Provide a direct contact between the oxidation and reduction electrodes - A</span>
Answer:
Option A
Explanation:
Remember constant means staying the same so in order to remain constant on the graph the line has to keep going up remaining the same pace. Which means the answer is option A or "Object A." The graph consists of a line going up at a constant speed without any bends or turns which all the other options consist of.
Hope this helps.
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)
Best regards.