From tables, the heat of combustion of ethanol (C₂H₅OH) is 29.7 kJ/g.
Given:
V = 400 mL of water
ΔT = 100 - 20 = 80°C, temperature rise
c = 1.00 cal/(g-°C), specific heat of water = 4.184 J/(g-°C)
Because the density of water is approximately 1.0 g/mL. the mass of water is
m = 400 g.
Let x = grams of ethanol that should be burned to make the water rise
by 80 °C.
Then
(29.7 x 10³ J/g)*(x g) = (400 g)*(4.184 J/g-°C))*(80 °C)
29.7 x 10³x = 1.339 x 10⁵
x = 4.5 g
Answer: 4.5 grams
Answer:
See below
Explanation:
Use Ideal Gas Law
PV = n RT using R = .082057366 l-atm/k-mol
T must be in Kelvin
solve for 'n'
.92 * 1.6 = n * .082057366 * 287
n = .0625 moles
then the mole weight: .0625 * x = .314
x = mole weight = 5.025 gm
"Systems with either very <u>low pressures or high temperatures</u> enable real gases to be estimated as “ideal.” "
Elements:
• atom
• cannot
• Periodic Table
Compounds:
• atoms
• chemically
• cannot
Mixtures:
• elements or compounds
• homogeneous
• heterogeneous
Answer:
1. Removing them to an area of fresh air. This helps to prevents further poisoning by the carbon monoxide and increase the amount of oxygen entering into the body. This will help to reduce the concentration of carbon monoxide binding oxygen
2. Administering pure oxygen goes a long way to enhance ventilation and increase the oxygen saturation to 100%. This will help to overcome the effect of the carbon monoxide and promote more hemoglobin binding
