Precipitation occurs when the product of the ion concentration exceeds the Ksp.
Answer:
<em>a)</em> <em>1.392 x 10^6 g/cm^3</em>
<em>b) 8.69 x 10^7 lb/ft^3</em>
<em></em>
Explanation:
mass of the star m = 2.0 x 10^36 kg
radius of the star (assumed to be spherical) r = 7.0 x 10^5 km = 7.0 x 10^8 m
The density of substance ρ = mass/volume
The volume of the star = volume of a sphere = 
==> V = 
= 1.437 x 10^27 m^3
density of the star ρ = (2.0 x 10^36)/(1.437 x 10^27) = 1.392 x 10^9 kg/m^3
in g/cm^3 = (1.392 x 10^9)/1000 = <em>1.392 x 10^6 g/cm^3</em>
in lb/ft^3 = (1.392 x 10^9)/16.018 = <em>8.69 x 10^7 lb/ft^3</em>
Data Given:
Pressure = P = ?
Volume = V = 3.0 L
Temperature = T = 115 °C + 273 = 388 K
Mass = m = 75.0 g
M.mass = M = 44 g/mol
Solution:
Let suppose the Gas is acting Ideally. Then According to Ideal Gas Equation,
P V = n R T
Solving for P,
P = n R T / V ------ (1)
Calculating Moles,
n = m / M
n = 75.0 g / 44 g.mol⁻¹
n = 1.704 mol
Putting Values in Eq. 1,
P = (1.704 mol × 0.08205 atm.L.mol⁻¹.K⁻¹ × 388 K) ÷ 3.0 L
P = 18.08 atm
Answer:
Alkanes, alkenes and alkynes are simple hydrocarbon chains with no functional groups. Alkanes are identified because the carbon chain has only single bonds. Alkenes have at least one double bond and alkynes have at least one triple bond. The most common alkyne is ethyne, better known as acetylene.
Answer:
125.5 ×10^-3 m^3= 0.1255 m^3
Explanation:
Volume=5.6mol×22.414dm^3
=125.5dm^3
