12) If a gas effuses 1.618 times faster than Kr (molar mass = 83.8 g/mol) what i
ID: 580538 • Letter: 1
Question
12) If a gas effuses 1.618 times faster than Kr (molar mass = 83.8 g/mol) what is the molar mass of the gas? (4 pts.) 13) For the following redox reaction, identify the oxidizing agent and the reducing agent. Cu(NO) (aq)+ Zn(s) 2Cu(s) Zn(NO,) (aq) ; Oxidizing agent Reducing agent (6 Pts.) In a titration experiment, 10.0 mL of 0.15 M Ba(OH)2 is required to titrate a 20.0 mL of an unknown solution of HNO. Calculate the concentration of the HNO, solution using the data. (6 pts.) 2HNO, Ba(OH Ba(NO,)2 +2H,O 14) 15) Given the following reaction: 21Cl, +3H,O BICI + HIO,+5HCI. If you begin with 25.0 g ICl, and 25.0 g H,O, how many grams of HIO, will be produced? (6 ptsExplanation / Answer
12) According to Graham's law of effusion:
Rate1/Rate2 = sqrt (M2/M1)
where M1, M2 are molar masses of the two gasses.
Here, we assume that the unknown gas is 1st gas and Krypton is 2nd gas.
Given that Rate1/Rate2 = 1.618, and moelcular weight of Krypton M2 = 83.8 g/mol.
Therefore,
1.618 = sqrt (M2/M1)
2.618 = (83.8 g/mol)/ M1
M1 = 32 g/mol
Therefore, the unknown gas molar mass = 32 g/mol
(this is probably oxygen diatomic gas)
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