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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Answer:
0.823 M was the molarity of the KOH solution.
Explanation:
(Neutralization reaction)
To calculate the concentration of base , we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is KOH.
We are given:
Putting values in above equation, we get:
0.823 M was the molarity of the KOH solution.
Answer:
25.7 kJ/mol
Explanation:
There are two heats involved.
heat of solution of NH₄NO₃ + heat from water = 0
q₁ + q₂ = 0
n = moles of NH₄NO₃ = 8.00 g NH₄NO₃ × 1 mol NH₄NO₃/80.0 g NH₄NO₃
∴ n = 0.100 mol NH₄NO₃
q₁ = n * ΔHsoln = 0.100 mol * ΔHsoln
m = mass of solution = 1000.0 g + 8.00 g = 1008.0 g
q₂ = mcΔT = 58.0 g × 4.184 J°C⁻¹ g⁻¹ × ((20.39-21)°C) = -2570.19 J
q₁ + q₂ = 0.100 mol ×ΔHsoln – 2570.19 J = 0
ΔHsoln = +2570.19 J /0.100 mol = +25702 J/mol = +25.7 kJ/mol
It has 4 significant figures. If you see the Zero after the one decimal point you don’t count that and instead just started at 6
Answer:
it's describes the velocity. since a direction was specifically given, that means it is displacement, and displacement is to velocity while distance is to speed
