This problem is looking for the minimum value of μs that is
necessary to achieve the record time. To solve this problem:
Assuming the front wheels are off the ground for the entire
¼ mile = 402.3 m, the acceleration a = µs·9.8 m/s².
For a constant acceleration, distance = 402.3
m = 1/2at^2 = 804.6 m / (4.43 s)^2 = a = µs·9.8 m/s^2
µs = 804.6 m / (4.43s)^2 / 9.8 m/s^2 = 4.18
So i converted everything first;
<span>7.0 C ---> 280 K </span>
<span>49 mL---> 0.049 L </span>
<span>74mL---> 0.074 L </span>
<span>THEN I tried setting it up by the combined law formula which is P1V1/T1=P2V2/T2 </span>
Answer: D)supersaturated
Explanation: Solubility is defined as the amount of solute in grams which can dissolve in 100 g of the liquid to form a saturated solution at that particular temperature.
At
, the solubility of
is 153g/100 ml.
Thus if 180 grams is dissolved, it contains more amount of solute than it can hold at that that temperature, and thus is supersaturated solution.
A saturated solution is a solution containing the maximum concentration of a solute dissolved in the solvent. The additional solute does not dissolve in a saturated solution.
An unsaturated solution is solution in which the solute concentration is lower than its equilibrium solubility.
A supersaturated solution is one that has more solute than it can hold at a certain temperature.
Answer:

Explanation:
As per thermal radiation we know that rate is heat radiation is given as

here we know that
T = 34 degree C = 307 K

e = 0.557


now we have



