Some
of the solutions exhibit
colligative properties. These properties depend on the amount of solute
dissolved in a solvent. These properties include freezing point depression, boiling
point elevation, osmotic pressure and vapor pressure lowering. Calculations
are as follows:
<span>
ΔT(freezing point) = (Kf)mi
3 = 1.86 °C kg / mol (m)(2)
3 =3.72m
m = 0.81 mol/kg</span>
Answer:
Gill's
Explanation:
Fish's have many characteristics that differ from humans, and one of them is the gills.
PH = -log [H+]
pH = -log (1.0x10^-4) = -(-4) = 4 or A
Answer:
- <u><em>1.7 × 10³ kg of ore.</em></u>
Explanation:
Call X the amount of aluminum ore mined to produce 1.0 × 10³ kg the aluminum metal.
Then, taking into account the yield of the reaction (82 % = 0.82) and the percent of aluminun in the ore (71% = 0.71), you can write the following equation:
- X × 71% × 82% = 1.0 × 10³ kg
↑ ↑ ↑ ↑
(mass of ore) (% of Al in the ore) (yield) ( Al metal to obtain)
You must just simplify, solve and compute:
- X = 1,000 / (0.71 × 0.82) = 1,000 / 0.5822 = 1,717.6 Kg
Round to two significant figures; 1,700 kg = 1.7 × 10³ kg of ore ← answer.
Answer:
The volume on the tank is 6, 20 L
Explanation:
We use the formula PV=nRT. We convert the units of pressure in kPa into atm and temperature in Celsius into Kelvin:
0°C=273K
101,325kPa---1 atm
275kPa --------x=(275kPax 1 atm)/101,325kPa= 2,71 atm
PV=nRT --> V=nRT/P
V= 0,750 mol x 0,082 l atm /K mol x 273 K/ 2, 71 atm= <em>6, 20 L</em>
