Answer:
Molecules along the surface of a liquid behave differently than those in the bulk liquid.
Cohesive forces attract the molecules of the liquid to one another.
Water forming a droplet as it falls from a faucet is a primary example of surface tension.
Explanation:
Surface tension is the force that stretches the liquid surface. This force acts normal to the surface. It is the downward force that acts on the surface of the liquids which is due to the cohesive forces of the liquids.
The water molecules are bonded by a strong hydrogen bond force which is between hydrogen atom and the electronegative oxygen atom. At the surface the water molecules are attracted strongly by other water molecules which lies below the surface and are stretched at the surface. Thus the water molecules at the surface acts differently than in the bulk liquid.
Mercury have a strong cohesive force than the water and have a higher surface tension force than the water.
Surface water acquires minimum surface area, hence acquiring spherical shape of water.
The Beer-Lambert law states that A = E*c*l where A is absorbance, E is the molar absorbance coeffecient, c is concentration and l is path length. Therefore the absorbance is directly proportional to concentration, and by increasing the concentration by a factor of 3, absorbance will increase by a factor of 3 giving A = 1.584
Answer:
At the burner temp. and pressure, 18.85 litres of air is needed to completely combust each gram of propane
Explanation:
The combustion stoichiometry is as follows:
C₃H₈ + 5O₂ = 4 H₂O + 3CO₂ The molecular weights (g/mol) are:
MW 44 5x32 4x18 3x44
So each gram of propane is 1/44 = 0.02272 mol propane
and will need 5 x 0.02272 = 0.1136 mol oxygen
At 0.21 mol fraction oxygen in air, 0.1136 / 0.21 = 0.54 mol air is needed to burn the propane.
At the low pressure in the burner we can use the Ideal Gas Law
PV=nRT, or V = nRT/P
P = 1.1 x 101325 Pa = 111457 Pa
T = 195°C + 273 = 468 K
R = 8.314
and we calculated n = number of moles air = 0.54 mol
So V m³ = 0.54 x 8.314 x 468 / 111457 = 0.0188 m³ = 18.85 litres air.
Answer:
B)
Explanation:
It is the theme of the passage.
Answer:
D = 28.2g
Explanation:
Initial temperature of metal (T1) = 155°C
Initial Temperature of calorimeter (T2) = 18.7°C
Final temperature of solution (T3) = 26.4°C
Specific heat capacity of water (C2) = 4.184J/g°C
Specific heat capacity of metal (C1) = 0.444J/g°C
Volume of water = 50.0mL
Assuming no heat loss
Heat energy lost by metal = heat energy gain by water + calorimeter
Heat energy (Q) = MC∇T
M = mass
C = specific heat capacity
∇T = change in temperature
Mass of metal = M1
Mass of water = M2
Density = mass / volume
Mass = density * volume
Density of water = 1g/mL
Mass(M2) = 1 * 50
Mass = 50g
Heat loss by the metal = heat gain by water + calorimeter
M1C1(T1 - T3) = M2C2(T3 - T2)
M1 * 0.444 * (155 - 26.4) = 50 * 4.184 * (26.4 - 18.7)
0.444M1 * 128.6 = 209.2 * 7.7
57.0984M1 = 1610.84
M1 = 1610.84 / 57.0984
M1 = 28.21g
The mass of the metal is 28.21g
