Answer:
If a car skids 66 ft on wet concrete, it will move at 243 ft/s when the brake is applied.
Explanation:
To determine how fast the car was moving, after skidding, the formula below is used:
V = √32*fd
V is the car's speed (ft/s)
d is skid length (ft) = 66 ft
f is the coefficient of friction determined by the material the car was skidding on.
Coefficient of friction for wet concrete is 0.65
V = √32*fd
V = √32 *0.65* 66
V = 242.679 ft/s ≅ 243 ft/s (nearest whole number)
If a car skids 66 ft on wet concrete, it will move at 243 ft/s when the brake is applied.
Answer:
A solid substance at its melting point has less energy than the same mass of the substance when it is a liquid at the same temperature. ... This heat energy allows the change of state to happen, and the temperature remains constant during the process.
Explanation:
First, balance the reaction:
_ KClO₃ ==> _ KCl + _ O₂
As is, there are 3 O's on the left and 2 O's on the right, so there needs to be a 2:3 ratio of KClO₃ to O₂. Then there are 2 K's and 2 Cl's among the reactants, so we have a 1:1 ratio of KClO₃ to KCl :
2 KClO₃ ==> 2 KCl + 3 O₂
Since we start with a known quantity of O₂, let's divide each coefficient by 3.
2/3 KClO₃ ==> 2/3 KCl + O₂
Next, look up the molar masses of each element involved:
• K: 39.0983 g/mol
• Cl: 35.453 g/mol
• O: 15.999 g/mol
Convert 10 g of O₂ to moles:
(10 g) / (31.998 g/mol) ≈ 0.31252 mol
The balanced reaction shows that we need 2/3 mol KClO₃ for every mole of O₂. So to produce 10 g of O₂, we need
(2/3 (mol KClO₃)/(mol O₂)) × (0.31252 mol O₂) ≈ 0.20835 mol KClO₃
KClO₃ has a total molar mass of about 122.549 g/mol. Then the reaction requires a mass of
(0.20835 mol) × (122.549 g/mol) ≈ 25.532 g
of KClO₃.
