The approximate volume of table tennis ball is 80 cm³
<h3>What is volume?</h3>
Volume is defined as the amount of space occupied by the three dimensional object. S I unit of volume is m³ or cm³.
To find the volume of tennis ball using graduated cylinder.
Step 1 - Fill the graduated cylinder half or full.
Step 2 - Mark the initial volume of the water i.e. 100 cm³ (Vi)
Step 3 - Put the tennis ball in the graduated cylinder. Some of the water was displaced by the table tennis ball.
Step 4 - Mark the Final volume of the water (Vf) i.e. 180 cm³
Step 5 = Calculate the volume by using Formula
Vb = Vf – Vi = 180 cm³ - 100 cm³ = 80 cm³
Hence the volume of tennis ball (Vb) is 80 cm³
For more Volume related question visit here:
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Destructive interference occurs when path difference = ½-integer
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<span>The periodic table is the most important chemistry reference there is. It arranges all the known elements in an informative array. Elements are arranged left to right and top to bottom in order of increasing atomic number. Order generally coincides with increasing atomic mass.
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Uncertainty means that your result may be very random, so you can't trust the first or second or so observation, making several samples critical for accuracy. <span />
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.
