Answer:
The molar mass of the gas is 44 g/mol
Explanation:
It is possible to solve this problem using Graham's law that says: Rates of effusion are inversely dependent on the square of the mass of each gas. That is:

If rate of effusion of nitrogen is Xdistance / 48s and for the unknown gas is X distance / 60s and mass of nitrogen gas is 28g/mol (N₂):

6,61 = √M₂
44g/mol = M₂
<em>The molar mass of the gas is 44 g/mol</em>
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I hope it helps!
Answer:
mendeleev left a space
Explanation:
so the periodic table can be organize
Answer:
Limited
Explanation:
For competition to exist within a population or between population, shared or common resources that organisms holds very important must be very limited and scared.
Competition is a struggle between organisms for limited resources in the ecosystem.
- It is an interaction between organisms in which one of them is harmed.
- Competition originates from limited supply of shared or mutual resources among organisms.
- When competition is between organisms from different population, it is called an interspecific competition.
- Competition between organisms within the same population, i.e of the same species is intraspecific competition.
Answer:
So first thing to do in these types of problems is write out your chemical reaction and balance it:
Mg + O2 --> MgO
Then you need to start thinking about moles of Magnesium for moles of Magnesium Oxide. Based on the above equation 1 mole of Magnesium is needed to make one mole of Magnesium Oxide.
To get moles of magnesium you need to take the grams you started with (.418) and convert to moles by dividing by molecular weight of Mg (24.305), this gives you .0172 moles of Mg.
The theoretical yield would be the assumption that 100% of the magnesium will be converted into Magnesium Oxide, so you would get, based on the first equation, .0172 mol of MgO. Multiplying this by the molecular weight of MgO (24.305+16) gives us .693 g of MgO.
The percent yield is what you actually got in the experiment, and for this you subtract off the total mass from the crucible mass, or 27.374 - 26.687, which gives .66 g of MgO obtained.
Percent yield is acutal/theoretical, .66/.693, or 95.24%.
I'll let you do the same for the second trial, and average percent yield is just an average of the two trials percent yield.
Hope this helps.