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.
A) Polymer is the general name of large units made of many smaller units (these would be called monomers). An example is starch, this is a carbohydrate polymer that is made up of smaller units (monomers) called glucose.
The container with chalk powder will contain the least amount of water, because it absorbs water, but the containers with honey and cocunut oil will conserve their amount of water, because they will prevent the water evaporation (especially cocunut oil because it will be on the top side of the container).
Molarity = number of moles of solute/liters of solution
number of moles of solute = molarity x liters of solution
Part (a): <span>30.00 ml of 0.100m Cacl2
number of moles of CaCl2 = 0.1 x 0.03 = 3x10^-3 moles
1 mole of CaCl2 contains 2 moles of chlorine, therefore 3x10^-3 moles of CaCl2 contains 6x10^-3 moles of chlorine
Part (b): </span><span>10.0 ml of 0.500m bacl2
number of moles of BaCl2 = 0.5 x 0.01 = 5x10^-3 moles
1 mole of BaCl2 contains 2 moles of chlorine, therefore 5x10^-3 moles of BaCl2 contains 10x10^-3 moles of chlorine
Part (c): </span><span>4.00 ml of 1.000m nacl
number of moles of NaCl = 1 x 0.004 = 0.004 moles
1 mole of NaCl contains 1 mole of chlorine, therefore 4x10^-3 moles of NaCl contains 4x10^-3 moles of chlorine
Part (d): </span><span>7.50 ml of 0.500m fecl3
number of moles of FeCl3 = 0.5 x 0.0075 = 3.75x10^-3 moles
1 mole of FeCl3 contains 3 moles of chlorine, therefore 3.75x10^-3 moles of FeCl3 contains 0.01125 moles of chlorine
Based on the above calculations, the correct answer is (d)</span>
Answer:
The value of the silver in the coin is 35.3 $
Explanation:
First of all, let's calculate the volume of the coin.
2π . r² . thickness = volume
r = diameter/2
r = 41 mm/2 = 20.5 mm
2 . π . (20.5 mm)² . 2.5 mm = 6601 mm³
Now, this is the volume of the coin, so we must find out how many grams are on it.
6601 mm³ / 1000 = 6.60 cm³
Let's apply density.
D = Mass / volume
10.5 g/cm³ = mass /6.60 cm³
10.5 g/cm³ . 6.60 cm³ = mass
69.3 g = mass
Each gram has a cost of 0.51$
69.3 g . 0.51$ = 35.3 $
