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.
Earthquakes and volcanoes most commonly occur around plate boundaries because of the movement from the plate boundaries. The interactions between the plates by moving under, upon, or sliding against other boundaries may cause earthquakes and volcanoes.
if they had a suitable amount to cause an interruption in the waves so huge and vast that it makes waves..... it depends because you can have any amount and get different results any day though
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Answer:
Concepts and Principles
1- Kinetic Energy: The kinetic energy of an object is:
K=1/2*m*v^2 (1)
where m is the object's mass and v is its speed relative to the chosen coordinate system.
2- Gravitational potential energy of a system consisting of Earth and any object is:
U_g = -Gm_E*m_o/r*E-o (2)
where m_E is the mass of Earth (5.97x 10^24 kg), m_o is the mass of the object, and G = 6.67 x 10^-11 N m^2/kg^2 is Newton's gravitational constant.
Solution
The argument:
My friend thinks that escape speed should be greater for more massive objects than for less massive objects because the gravitational pull on a more massive object is greater than the gravitational pull for a less massive object and therefore the more massive object needs more speed to escape this gravitational pull.
The counterargument:
We provide a mathematical counterargument. Consider a projectile of mass m, leaving the surface of a planet with escape speed v. The projectile has a kinetic energy K given by Equation (1):
K=1/2*m*v^2 (1)
and a gravitational potential energy Ug given by Equation (2):
Ug = -G*Mm/R
where M is the mass of the planet and R is its radius. When the projectile reaches infinity, it stops and thus has no kinetic energy. It also has no potential energy because an infinite separation between two bodies is our zero-potential-energy configuration. Therefore, its total energy at infinity is zero. Applying the principle of energy consersation, we see that the total energy at the planet's surface must also have been zero:
K+U=0
1/2*m*v^2 + (-G*Mm/R) = 0
1/2*m*v^2 = G*Mm/R
1/2*v^2 = G*M/R
solving for v we get
v = √2G*M/R
so we see v does not depend on the mass of the projectile
