Answer:
0.482 ×10²³ molecules
Explanation:
Given data:
Volume of gas = 2.5 L
Temperature of gas = 50°C (50+273 = 323 k)
Pressure of gas = 650 mmHg (650/760 =0.86 atm)
Molecules of N₂= ?
Solution:
PV= nRT
n = PV/RT
n = 0.86 atm × 2.5 L /0.0821 atm. mol⁻¹. k⁻¹. L × 323 k
n = 2.15 atm. L /26.52 atm. mol⁻¹.L
n = 0.08 mol
Number of moles of N₂ are 0.08 mol.
Number of molecules:
one mole = 6.022 ×10²³ molecules
0.08×6.022 ×10²³ = 0.482 ×10²³ molecules
Land will warm faster/quicker
Answer: 362,07 cm3
To answer this question you need to convert the lb into gram first. One lb equal to 453.592g, so: 3.6lb x 453.592gram/lb= 1632.9312gram.
Now we have mass(1632.9312g), density (4.51g/cm3). Volume is mass divided by density. The equation would be:
Volume= mass/density
Volume = 1632.9312gram / (4.51g/cm3)= 362,07 cm3
The mass of ammonium chloride that must be added is : ( A ) 4.7 g
<u>Given data :</u>
Volume of water ( V ) = 250 mL = 0.25 L
pH of solution = 4.85
Kb = 1.8 * 10⁻⁵
Kw = 10⁻¹⁴
Given that the dissolution of NH₄Cl gives NH₄⁺⁺ and Cl⁻ ions the equation is written as :
NH₄CI + H₂O ⇄ NH₃ + H₃O⁺
where conc of H₃O⁺
[ H₃O⁺ ] =
and Ka = Kw / Kb
∴ Ka = 5.56 * 10⁻¹⁰
Next step : Determine the concentration of H₃O⁺ in the solution
pH = - log [ H₃O⁺ ] = 4.85
∴ [ H₃O⁺ ] in the solution = 1.14125 * 10⁻⁵
Next step : Determine the concentration of NH₄CI in the solution
C = [ H₃O⁺ ]² / Ka
= ( 1.14125 * 10⁻⁵ )² / 5.56 * 10⁻¹⁰
= 0.359 mol / L
Determine the number of moles of NH₄CI in the solution
n = C . V
= 0.359 mol / L * 0.25 L = 0.08979 mole
Final step : determine the mass of ammonium chloride that must be added to 250 mL
mass = n * molar mass
= 0.08979 * 53.5 g/mol
= 4.80 g ≈ 4.7 grams
Therefore we can conclude that the mass of ammonium chloride that must be added is 4.7 g
Learn more about ammonium chloride : brainly.com/question/13050932