Answer:
The correct answer is option 2.
Explanation:
Colligative property is defined as property which depends upon only on the numbers of particles of solute dissolved in definite amount of solvent, It do not depend on the nature of the solute.
For example : NaCl solution with 0.4 molal will show same colligative properties as a that of the glucose solution with 0.04 molal concentration.
The following are the examples of colligative property:
1. Relative lowering of vapor pressure.
2. Osmotic pressure
3. Elevation in boiling points
4. Depression in freezing point
The answer is C, because wind turbines are used to produce electricity, and it's also the most safest way--but they are expensive.
Answer:
The new volume is 5.913*10^4 L
Explanation:
Step 1: Write out the formula to be used:
Using general gas equation;
P1V1 / T1 =P2V2 /T2
V2 = P1V1T2 / P2T1
Step 2: write out the values given and convert to standard unit's where necessary
P1 = 0.995atm
P2 0.720atm
V1 = 5*10^4 L
T1 = 32°C = 32+ 273 = 305K
T2 = -12°C = -12 + 273 = 261K
Step 3: Equate your values and do the calculation:
V2 = 0.995 * 5*10^4 * 261 / 0.720 * 305
V2 = 1298.475 * 10^4 / 219.6
V2 = 5.913 * 10^4 L
So the new volume of the balloon is 5.913*10^4 L
The combustion reaction of propane would be expressed as:
C3H8 + 5O2 = 3CO2 + 4H2O
To determine the mass of water that is produced from the given amount of propane, we use the mass of propane and the relation of the substances from the balanced reaction. We do as follows:
moles propane = 22 g C3H8 ( 1 mol / 44.1 g ) = 0.50 mol C3H8
moles H2O = 0.50 mol C3H8 ( 4 mol H2O / 1 mol C3H8) = 2 mol H2O
mass H2O = 2 mol H2O ( 18.02 g / 1 mol ) = 36.04 g H2O
Therefore, the mass of water that is produced from 22 grams of propane would be 36.04 g.
