Answer:
The empirical formula would be N₂Os * Page 2 Calculate the empirical formulaof a compound that is 94.1% oxygen, 5.9% hydrogen.
Answer: State
Explanation:
The rock on the left is more shiny than the right, so that’s incorrect.
The rocks are not the same color, so that’s also incorrect.
Even though you can’t feel the textures, the right rock looks grainy and the left on doesn’t.
They both are solid, so that is the correct answer.
Burning gas is a chemical change and evaporation of gas is physical change.
Answer:
0.4076 g
Explanation:
Kp is the equilibrium constant based on pressure and depends only on gas substances. For a generic reaction
aA + bB ⇄ cC + dD
, where pX is the pressure of X in equilibrium.
For the reaction Kp = pCO₂
pCO₂ = 0.026 atm
The system is in equilibrium at the beginning. The compression occurs at a constant temperature, so using Boyle's law
P1V1 = P2V2
0.026*10 = P2*0.1
P2 = 2.6 atm
The reaction will reach again the equilibrium, and pCO₂ = 0.026 atm, then the rest will form MgCO₃, which will be 2.6 - 0.026 = 2.574 atm.
By the ideal gas law:
PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.082 atm*L/mol*K), and T is the temperature.
2.574*0.1 = n*0.082*650
53.3n = 0.2574
n = 4.83x10⁻³ mol
The stoichiometry of the reaction is 1 mol of MgCO₃ for 1 mol of CO₂, so it will form 4.83x10⁻³ mol of MgCO₃ .
The molar mass is:
MgCO₃: 24 g/mol of Mg + 12 g/mol of C + 3*16 g/mol of O = 84 g/mol
The mass formed is the molar mass multiplied by the number of moles:
m = 84x4.83x10⁻³
m = 0.4076 g