<u>We are given:</u>
The force applied on the poor hamster (F) = 12 N
Acceleration of the poor Hamster (a) = 8 m/s²
<u>Solving for the mass of the Poor Hamster:</u>
From newton's second equation of motion, we know that:
F = ma
<em>replacing the given values</em>
12 = 8 * m
m = 12/8 kg
m = 3/2 kg
The poor Hamster weighs 3/2 kg
Answer:
- NaClO₃ > KBr > KNO₃ > NaCl.
Explanation:
The attached file contains the graph with the solubility curves for the four substances, KNO₃, NaClO₃, KBr, NaCl.
To determine the solubility of each salt at a certain temperature, you read the temperature on the horizontal axis, labeled Temperature (ºC), and move upward up to intersecting the curve of the corresponding salt. Then, move horizontally up to insersceting the vertical axis, labeled Solubility (g/100g of H₂O), to read the solubility.
The higher the reading on the vertical axis, the higher the solubility.
The red vertical line that I added is at a temperature of 40ºC.
The number in blue indicate the order in which the solubility curves are intersected at that temperature:
- 4: NaCl: this is the lowest solubility
- 3: KNO₃: this is the second lowest solubility
- 2: KBr: this is the third lowest solubility
- 1: NaClO₃: this is the highest solubility.
Thus, the rank, from most soluble to least soluble is:
- NaClO₃ > KBr > KNO₃ > NaCl.
answer: 3g. 17kg+3 ÷ 0.25
Answer:
333.7g of antifreeze
Explanation:
Freezing point depression in a solvent (In this case, water) occurs by the addition of a solute. The law is:
ΔT = Kf × m × i
Where:
ΔT is change in temperature (0°C - -20°C = 20°C)
Kf is freezing point depression constant (1.86°C / m)
m is molality of solution (moles solute / 0.5 kg solvent -500g water-)
i is Van't Hoff factor (1, assuming antifreeze is ethylene glycol -C₂H₄(OH)₂)
Replacing:
20°C = 1.86°C / m × moles solute / 0.5 kg solvent × 1
5.376 = moles solute
As molar mass of ethylene glycol is 62.07g/mol:
5.376 moles × (62.07g / 1mol) = <em>333.7g of antifreeze</em>.
Answer:
different cell specialized in different function