The mass of nitrogen collected is mathematically given as
M-N2=0.025gram
<h3>What is the mass of nitrogen collected?</h3>
Question Parameters:
A sample weighing 2.000g
the liberated NH3 is caught in 50ml pipeful of H2SO4 (1.000ml = 0.01860g Na2O).
T=26.3c=299.3K
Pressure=745mmHg=745torr
Pressure of N2=745-25.2=719.8torr
Generally, the equation for the ideal gas is mathematically given as
PV=nRT
Therefore
719.8/760=45.6/1000=n*0.0821*299.3
n=0.00176*14
In conclusion, the Mass of N2
M-N2=0.00176*14
M-N2=0.025gram
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When 1.00 L of a 0.500 M LiBr solution is tripled by dilution with water . the number of moles of lithium in new solution is 0.5 moles.
given that :
volume = 1.00 L
molarity = 0.500 M
The solution is tripled by dilution but there will be no effect in number of mole by the dilution.
the number of moles = molarity × volume in L
the number of moles = 1 × 0.500
the number of moles = 0.5 mol
Thus, When 1.00 L of a 0.500 M LiBr solution is tripled by dilution with water . the number of moles of lithium in new solution is 0.5 moles.
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Answer:
Trial Number of moles
1 0.001249mol
2 0.001232mol
3 0.001187 mol
Explanation:
To calculate the <em>number of moles of tritant</em> you need its<em> molarity</em>.
Since the<em> molarity</em> is not reported, I will use 0.1000M (four significant figures), which is used in other similar problems.
<em>Molarity</em> is the concentration of the solution in number of moles of solute per liter of solution.
In this case the solute is <em>NaOH</em>.
The formula is:
Solve for the <em>number of moles:</em>
Then, using the molarity of 0.1000M and the volumes for each trial you can calculate the number of moles of tritant.
Trial mL liters Number of moles
1 12.49 0.01249 0.01249liters × 0.1000M = 0.001249mol
2 12.32 0.01232 0.01232liters × 0.1000M = 0.001232mol
3 11.87 0.01187 0.01187liters × 0.1000M = 0.001187 mol
