Answer:
N2
Explanation:
Rate of effusion is defined by Graham's Law:
(Rate 1/Rate 2) = (sqrt (M2)/ sqrt (M1))
(Where M is the molar mass of each substance. )
Molar Mass of oxygen, O2, is 32 (M1).
Rate of effusion of O2 to an unknown gas is .935(Rate 1).
Rate 2 is unknown so put 1.
Solve for x (M2).
.935/1 = sqrt x/ sqrt32
.935 x sqrt 32 = sqrt x
5.29 = sq rt x
5.29^2 = 27.975 = 28
N2 has a molar mass of 28 so it is the correct gas.
Answer:
D
Explanation:
I believe the answer is D.
Answer:
Purpose: To become familiar with the techniques for separation of amixture of solids.
Explanation:
a mixture of pure substances. If you have a mixture of tennis ballsand marbles (not pure substances by the way), it would be easy toseparate the mixture. However, it is more difficult to separate asand (also not a pure substance) and salt mixture. Even with verygood tweezers and a magnifying glass, it would be extremelytedious. You could take advantage of the fact that salt dissolvesin water and sand does not. To separate iron powder from an ironand sand mixture you can take advantage of the magnetic propertiesof iron and separate the mixture.
To summarize a complete procedure for separating a mixture ofseveral substances, it is best to prepare a flow chart. A flowchartis a schematic representation of an algorithm or a stepwiseprocess, showing the steps as boxes of various kinds, and theirorder by connecting these with arrows. Flowcharts are used indesigning or documenting a process.
Hello there,
Ammonia will have a total of 4 atoms.
Hope I helped :)
The balanced chemical reaction would be as follows:
<span>5P4O6 +8I2 ---> 4P2I4 +3P4O10
We are given the amount of reactants used for the reaction. We first need to determine the limiting reactant from the given amounts. We do as follows:
8.80 g P4O6 (1 mol / </span><span>219.88 g) = 0.04 mol P4O6
12.37 g I2 ( 1 mol / </span><span>253.809 g ) = 0.05 mol I2
Therefore, the limiting reactant is iodine since less it is being consumed completely in the reaction. We calculate the amount of P2I4 prepared as follows:
0.05 mol I2 ( 4 mol P2I4 / 8 mol I2 ) (</span><span>569.57 g / 1 mol) = 14.24 g P2I4</span>
