Some guidance notes which may help.To calculate the current flow, Ohm's law can be used. This can be written as current=voltage/resistance, or I=V/R. V is 1.5V.R for the copper wire quoted would be calculated as R = resistivity x length/cross sectional area. The area would be calculated from the formula area = pi x diameter squared/4So, R=resistivity x length divided by (pi x diameter squared/4)Until is the resistivity of copper is known, that's about as far as can be gone.Any further questions, please ask.
Answer:
v = 20.31 m/s
Explanation:
p = mv -> v = p/m = 32,500 kg*m/s / 1,600 kg = 20.31 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.
Answer:
5 hours
Explanation:
formula is 6 km per hour
if you have to travel 30 km, divide 30 by 6
30/6 = 5 hours
