Answer: beta, gamma, alpha
Explanation: Beta is weakest gamma is middle alpha is strongest
Volume of Hydrogen V1 = 351mL
Temperature T1 = 20 = 20 + 273 = 293 K
Temperature T2 = 38 = 38 + 273 = 311 K
We have V1 x T2 = V2 x T1
So V2 = (V1 x T2) / T1 = (351 x 311) / 293 = 372.56
Volume at 38 C = 373 ml
<u>Answer:</u> The solubility of 
in water is 
<u>Explanation:</u>
The balanced equilibrium reaction for the ionization of cadmium phosphate follows:
3s 2s
The expression for solubility constant for this reaction will be:
![K_{sp}=[Cd^{2+}]^3[PO_4^{3-}]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCd%5E%7B2%2B%7D%5D%5E3%5BPO_4%5E%7B3-%7D%5D%5E2)
We are given:
Putting values in above equation, we get:
Hence, the solubility of in water is
Answer:
Amount of excess Carbon (ii) oxide left over = 23.75 g
Explanation:
Equation of the reaction: Fe₂O₃ + 3CO ----> 2Fe + 3CO₂
Molar mass of Fe₂O₃ = 160 g/mol;
Molar mass of Carbon (ii) oxide = 28 g/mol
From the equation of reaction, 1 mole of Fe₂O₃ reacts with 3 moles of carbon (ii) oxide; i.e. 160 g of iron (iii) oxide reacts with 84 g (3 * 28 g) of carbon (ii) oxide
450 g of Fe₂O₃ will react with 450 * 84/180) g of carbon (ii) oxide = 236..25 g of carbon (ii) oxide
Therefore the excess reactant is carbon (ii) oxide.
Amount of excess Carbon (ii) oxide left over = 260 - 236.25
Amount of excess Carbon (ii) oxide left over = 23.75 g
BAS might be an improperly capitalized: BAs, BaS
PTF2 might be an improperly capitalized: PtF2
BAF2 might be an improperly capitalized: BaF2
PTS might be an improperly capitalized: PtS
