<h3>Answer;</h3>
<em><u>Sand Spit or Spit </u></em>
<h3><u>Explanation;</u></h3>
- <em><u>Long shore drift is the process that occurs when a sheet of water moves on and off the beach, in other words the swash and back swash</u></em>, thus capturing and transporting sediment on the beach back out to the sea.
- <em><u>Sandbar</u></em> is normally formed when the sandspit stretches across a bay and connects the two sides. <em><u>Headland</u></em> is a high piece of land that extends out onto the sea. <em><u>Sea stacks </u></em>on the other hand results from the collapsing of the roof of the arch.
Answer:
acceleration = v-u /t
30- 20/5
= 10/5 = 2m/sec²
Force = mass * acceleration
Force = 0.1 * 2
Force = 0.2 Newton
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.
