I know what you're asking but I don't think the question is stated properly. Technically, an atom will not join with an "oxide" ion; i.e., the oxide ion is an atom of oxygen to which two electrons have been added. An oxide ion will add to 2 K ions or 1 Ca ion. The K ion has lost just one electron so it takes two of them to equal the 2- charge on the oxide ion whereas the Ca ion has lost two electrons and it takes only one of them to equal the charge on the oxide ion.
Answer:
6.05g
Explanation:
The reaction is given as;
Ethane + oxygen --> Carbon dioxide + water
2C2H6 + 7O2 --> 4CO2 + 6H2O
From the reaction above;
2 mol of ethane reacts with 7 mol of oxygen.
To proceed, we have to obtain the limiting reagent,
2,71g of ethane;
Number of moles = Mass / molar mass = 2.71 / 30 = 0.0903 mol
3.8g of oxygen;
Number of moles = Mass / molar mass = 3.8 / 16 = 0.2375 mol
If 0.0903 moles of ethane was used, it would require;
2 = 7
0.0903 = x
x = 0.31605 mol of oxygen needed
This means that oxygen is our limiting reagent.
From the reaction,
7 mol of oxygen yields 4 mol of carbon dioxide
0.2375 yields x?
7 = 4
0.2375 = x
x = 0.1357
Mass = Number of moles * Molar mass = 0.1357 * 44 = 6.05g
Answer:
7.5 M
Explanation:
In order to find a solution's molar concentration, or molarity, you need to determine how many moles of solute, which in your case is sodium sulfate,
Na
2
SO
4
, you get in one liter of solution.
That is how molarity was defined -- the number of moles of solute in one liter of solution.
So, you know that you have
0.090
moles of solute in
12 mL
of solution. Your goal here will be to scale up this solution by using this information as a conversion factor to help you determine the number of moles of solute present in
According to
Graham's Law ," the rates of effusion or diffusion of two gases are inversely proportional to the square root of their molecular masses at given pressure and temperature".
r₁ / r₂ =
---- (1)
r₁ = Rate of effusion of He
r₂ = Rate of Effusion of O₃
M₁ = Molecular Mass of He = 4 g/mol
M₂ = Molecular Mass of O₃ = 48 g/mol
Putting values in eq. 1,
r₁ / r₂ =
r₁ / r₂ =
r₁ / r₂ =
3.46
Result: Therefore, Helium will effuse
3.46 times more faster than Ozone.
Answer:
387 g/mol
Explanation:
The molar mass is a ratio comparing a substance's mass and molar value. The specific ratio looks like this:
Molar Mass (g/mol) = mass (g) / moles
You can plug the given values into the ratio to find the molar mass.
Molar Mass = mass / moles
Molar Mass = 0.406 g / 0.00105 mol
Molar Mass = 387 g/mol
