The average speed of the whole travel is equal to <u>400 mph</u>.
Why?
From the statement, we know that whole travel is divided into three parts. For the first part (traveling from New York to Chicago), we have that it was 3.25 hours and the covered distance was half of the total distance (1400mi). For the second part, we have that it was 1 hour (layover time), and the covered no distance. For the third part (traveling from Chicago to Los Angeles), we have that it was 2.75 hours, and it took the other half of the total distance (1400mi).
We can calculate the average speed of the whol travel using the following formula:
Now, substituting and calculating, we have:
Hence, we have the average speed of the whole travel is equal to 400 mph.
Have a nice day!
Answer:
2.11 m/s
Explanation:
Take north to be positive and south to be negative.
Average velocity = displacement / time
v = (82 m + -44 m) / (14 s + 4 s)
v = 2.11 m/s
The velocity is positive, so it is 2.11 m/s north. The magnitude of the velocity is 2.11 m/s.
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.
