Empirical formula is calculated as follows
calculate the moles of each element, that is % composition/ molar mass
molar masses ( Si= 28.09g/mol , Cl= 35.5 g/mol, I=126.9 g/mol)
moles of silicon = 7.962/28.09g/mol= 0.283 moles
moles of chlorine = 20.10 / 35.5g/mol = 0.566 moles
moles of iodine= 71.94 / 126.9 g/mol= 0.567 moles
divide each mole with smallest mole (0.283)
that is silicon = 0.283/0.283= 1 mole
chlorine = 0.566/0.283= 2 mole
Iodine= o.567/0.283= 2 moles
empirical formula is therefore= SiCl2I2
When the salt AgI dissolves, it dissociates as follows;
AgI --> Ag⁺ + I⁻
molar solubility of salt is the amount of salt that can be dissolved in 1 L of solution
since the ions dissociated are in 1:1 molar ratio, the molar solubility of the ions are equivalent to the molar solubility of the salt.
ksp is the solubility product constant of the salt
ksp = [Ag⁺][I⁻]
ksp = (9.1 x 10⁻⁹ mol/L)²
ksp = 8.28 x 10⁻¹⁷
Answer: C(s) + O2(g) --> CO2(g)12g (C) .... 50.8g (O2)................. initial amounts0g(C) .........18.8g(O2) ................. amounts when reaction completeThat means that C was the limiting reactant, and the amount of CO2 is based on the amount of carbon that burned. Covert 12 grams of carbon to moles. The moles of CO2 will be the same, since they are in a 1:1 mole ratio. Then convert the moles of CO2 to grams.12g C x (1 mol C / 12.0 g C) x (1 mol CO2 / 1 mol C) x (44.0g CO2 / 1 mol CO2) =44 g of CO2
A model is a purposeful representation of reality. When law is more complex and dynamic. Law is the description of an observed phenomenon. Make sense?
Hopefully that helped :)
Answer:
B. CH3COOH pH > 4.7 (4.8)
Explanation:
- CH3COOH + NaOH ↔ CH3COONa + H2O
- CH3COONa + NaOH ↔ CH3COONa
∴ mol NaOH = (5 E-3 L)*(0.10 mol/L) = 5 E-4 mol
⇒ mol CH3COOH = (0.05 L)*(0.20 mol/L) = 0.01 mol
⇒ <em>C</em> CH3COOH = (0.01 mol - 5 E-4 mol) / (0.105 L)
⇒ <em>C</em> CH3COOH = 0.0905 M
∴ mol CH3COONa = (0.05 L )*(0.20 mol/L) = 0.01 mol
⇒ <em>C</em> CH3COONa = (0.01 mol + 5 E-4 mol) / (0.105 L )
⇒ <em>C</em> CH3COONa = 0.1 M
∴ Ka = ([H3O+]*(0.1 + [H3O+])) / (0.0905 - [H3O+]) = 1.75 E-5
⇒ 0.1[H3O+] + [H3O+]² = (1.75 E-5)*(0.0905 - [H3O+])
⇒ [H3O+]² 0.1[H3O+] = 1.584 E-6 - 1.75 E-5[H3O+]
⇒ [H3O+]² + 0.1000175[H3O+] - 1.584 E-6 = 0
⇒ [H3O+] = 1.5835 E-5 M
∴ pH = - Log [H3O+]
⇒ pH = - Log (1.5835 E-5)
⇒ pH = 4.8004 > 4.7
