Answer:
X = Water (H2O) ; Y = Hydrogen ; Z = Oxygen
Explanation:
2(H2O) -------> 2H2 + O2
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.
Answer:
THE VOLUME OF THE BALLOON IS 1.45 L
Explanation:
At sea level:
Volume = 2 L
Pressure = 1 atm
Temperature = 12 °C
At 30000 ft altitude:
Pressure = 0.30 atm
Temperature = -55°C
Volume = unknown
Using the general gas formula:
P1 V1 / T1 = P2 V2 / T2
Re-arranging the formula by making V2 the subject of the equation, we have;
V2 = P1 V1 T2 / T1 P2
V2 = 1 * 2 * 12 / 0.30 * 55
V2 = 24 / 16.5
V2 = 1.45 L
The volume of the balloon at the temperature of -55 C and 0.30 atm is 1.45 L
Answer:
Percentage mass of copper in the sample = 32%
Explanation:
Equation of the reaction producing Cu(NO₃) is given below:
Cu(s)+ 4HNO₃(aq) ---> Cu(NO₃)(aq) + 2NO₂(g) + 2H₂O(l)
From the equation of reaction, 1 mole of Cu(NO₃) is produced from 1 mole of copper. Therefore, 0.010 moles of Cu(NO₃) will be produced from 0.010 mole of copper.
Molar mass of copper = 64 g/mol
mass of copper = number of moles * molar mass
mass of copper = 0.01 mol * 64 g/mol = 0.64 g
Percentage by mass of copper in the 2.00 g sample = (0.64/2.00) * 100%
Percentage mass of copper in the sample = 32%
