If the.pressure exerted by a gas at [math]25^{\circ} \mathrm{C}[/math] in a volume of 0.044 L is 3.81 atm, how many moles of gas are present
KE=3070.625 J
Height = 3.686 m
<h3>Further explanation</h3>
mass of bike+rider=85 kg
velocity = 8.5 m/s
Conservation of energy :
(KE+PE)₁ (downhill) = (KE+PE)₂ (up the hill)
PE₁=0⇒h=0
KE₂=0⇒v=0(stop), so equation becomes :
KE₁=PE₂
Answer:
60%
Explanation:
M(NH4NO3) = 2*14 +4*1 +3*16 = 80 g/mol
M(3O) = 3*16 = 48 g/mol
(48/80) *100 % =60% oxygen by mass.
Answer:
Enthalpy of vaporization = 30.8 kj/mol
Explanation:
Given data:
Mass of benzene = 95.0 g
Heat evolved = 37.5 KJ
Enthalpy of vaporization = ?
Solution:
Molar mass of benzene = 78 g/mol
Number of moles = mass/ molar mass
Number of moles = 95 g/ 78 g/mol
Number of moles = 1.218 mol
Enthalpy of vaporization = 37.5 KJ/1.218 mol
Enthalpy of vaporization = 30.8 kj/mol
Answer:
We need 92.3 grams of sodium azide
Explanation:
Step 1: Data given
Mass of nitrogen gas = 59.6 grams
Molar mass of nitrogen gas = 28.0 g/mol
Molar mass of sodium azide = 65.0 g/mol
Step 2: The balanced equation
2NaN3 → 2Na + 3N2
Step 3: Calculate moles nitrogen gas
Moles N2 = mass N2 / molar mass N2
Moles N2 = 59.6 grams/ 28.0 g/mol
Moles N2 = 2.13 moles
Step 4: Calculate moles NaN3
for 2 moles NaN3 we'll have 2 moles Na and 3 moles N2
For 2.13 moles N2 we need 2/3* 2.13 = 1.42 moles NaN3
Step 5: Calculate mass NaN3
Mass NaN3 = Moles NaN3 * molar mass NaN3
Mass NaN3 = 1.42 moles * 65.0 g/mol
Mass NaN3 = 92.3 grams
We need 92.3 grams of sodium azide
