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SOLUTIONS
↪1) Aqueous Solution
↪2) Solvent
↪3) Solute
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Use the Ideal Gas Law to find the moles of gas first.
Be sure to convert T from Celsius to Kelvin by adding 273.
Also I prefer to deal with pressure in atm rather than mmHg, so divide the pressure by 760 to get it in atm.
PV = nRT —> n = PV/RT
P = 547 mmHg = 547/760 atm = 0.720 atm
V = 1.90 L
T = 33°C = 33 + 273 K = 306 K
R = 0.08206 L atm / mol K
n = (0.720 atm)(1.90 L) / (0.08206 L atm / mol K)(306 K) = 0.0545 mol of gas
Now divide grams by mol to get the molecular weight.
3.42 g / 0.0545 mol = 62.8 g/mol
Answer:
0.032 L or 32 mL
Explanation:
Use the dilution equation M1V1 = M2V2
M1 = 9.0 M
V1 = This is what we're looking for.
M2 = 0.145 M
V2 = 2 L
Solve for V1 --> V1 = M2V2/M1
V1 = (0.145 M)(2 L) / (9.0 M) = 0.032 L
Answer:
it will a i did the quiz got it all right
Explanation:
Answer:
The acid must be a concentrated acid
Explanation:
Ethene is prepared in the laboratory by heating ethanol with excess concentrated tetraoxosulphate VI acid at 170°C . The reactionoccursc in two stages;
1) when the ethanol and sulphuric acid are mixed in a ratio of 1:2, ethyl hydrogentetraoxosulpate VI is formed
2) The compound formed in the first step is heated in the presence of excess concentrated sulphuric acid to give ethene and sulphuric acid.
The overall reaction can be perceived as the dehydration of ethanol. The gas produced (ethene) is usually passed through sodium hydroxide solution to remove any gaseous impurities present.
concentrated sulphuric acid is used in this process since it is a good dehydrating agent.
