Answer:
See Explanation
Explanation:
These problems are based upon the Ideal Gas Law, PV = nRT. Variables are gas phase variables and include ...
Pressure (P)
Volume (V)
Mass (moles (n))
R = Gas Constant = 0.08206L·atm/mol·K
Temperature (T)
Ideal Gas Law => PV = nRT
Note => when using the gas constant, R, convert all dimensional units to match those of L·atm/mol·K.
Problem 1 => Given ...
P = nRT/V
V = 250 ml = 0.250 L
n = 3.2 moles
R = 0.08206L·atm/mol·K
T = 45°C = (45 + 273)K = 318K
P = nRT/V = (3.2mol)(0.08206L·atm/mol·K)(318K)/(0.250L) = 334atm = 334atm(760mmHg/atm) = 253,853mmHg ≅2.5 x 10⁵mmHg (2 sig. figs.)
Problem 2 = Given ...
P = 5.05atm
V = nRT/P
n = 4.25 moles
R = 0.08206L·atm/mol·K
T = 27°C = (27 + 273)K = 300K
V = (4.25mol)(0.08206L·atm/mol·K)(300K)/(5.05atm) = 20.7 Liters O₂(g)
Problem 3 = Given ...
P = 302 KPa = (302KPa)(0.01atm/KPa) = 3.02 atm
V = 4.5 Liters
n = PV/RT
R = 0.08206L·atm/mol·K
T = 15°C = (15 + 273)K = 288K
n = (3.02atm)(4.5L)/(0.08206L·atm/mol·K)(288k) = 0.58 mole O₂ = 0.58 mole x 32 g/mole = 18.4 grams O₂
Problem 4 = Given ...
P = 810mmHg = 810mmHg(1atm/760mmHg) = 1.07atm
V = 2.0 Liters
n = 0.56 mole N₂(g)
R = 0.08206L·atm/mol·K
T = PV/nR = (1.07atm)(2.0L)/(0.56mol)(0.08206L·atm/mol·K) = 46.6K = (46.6 - 273)°C = -226°C
Problem 5 = Given ...
P = 3.2 atm
V = 55 Liters
n = PV/RT
R = 0.08206L·atm/mol·K
T = 17°C = (17 + 273)K = 290K
n = (3.2atm)(55L)/(0.08206L·atm/mol·K)(290K) = 7.4 moles N₂(g)
Always pay attention even the littlest thing can mess everything up
Explanation:
One is amplitude, which is the distance from the rest position of a wave to the top or bottom. Large amplitude waves contain more energy. The other is frequency, which is the number of waves that pass by each second. If more waves pass by, more energy is transferred each second.
<h3>
Answer:</h3><h3>a) 9.033 × 10²³ particles</h3><h3>b) 4.068 × 10²⁴ particles</h3><h3>c) 1.51 × 10²³ particles</h3>
Explanation:
For us to answer these questions, we have to know two formulas:
- Number of particles = moles × Avogadro's Number
- Moles = Mass ÷ Molar Mass
Therefore:
a) particles of Na = 1.50 mol × (6.022 × 10²³) particles/mol
= 9.033 × 10²³ particles
b) particles of Pb = 6.755 mol × (6.022 × 10²³) particles/mol
= 4.068 × 10²⁴ particles
c) particles of Si
= (7.02 g ÷ 28.085 g/mol) × (6.022 × 10²³) particles/mol
= 1.51 × 10²³ particles
Answer:
The correct answer is b) 2
Explanation:
When is dissolved in water, silver acetate (AgCH₃COO) is dissociated into ions according to the following equilibrium:
AgCH₃COO ⇄ Ag⁺ + CH₃COO⁻
Where Ag⁺ is a silver cation and CH₃COO⁻ is the acetate anion (an organic anion). As we can see, from one single formula unit are obtained 2 ions (1 cation and 1 anion).
Therefore, the correct option is b) - 2