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:
An open umbrella would not hit the ground in front of the closed one.
Explanation:
The closed umbrella would impact first because it did not have as much air pressure. The open one has air that presses against it and makes it reach where the open umbrella floats but falls slowly
Answer:
g(h) = g ( 1 - 2(h/R) )
<em>*At first order on h/R*</em>
Explanation:
Hi!
We can derive this expression for distances h small compared to the earth's radius R.
In order to do this, we must expand the newton's law of universal gravitation around r=R
Remember that this law is:
In the present case m1 will be the mass of the earth.
Additionally, if we remember Newton's second law for the mass m2 (with m2 constant):
Therefore, we can see that
With a the acceleration due to the earth's mass.
Now, the taylor series is going to be (at first order in h/R):
a(R) is actually the constant acceleration at sea level
and
Therefore:
Consider that the error in this expresion is quadratic in (h/R), and to consider quadratic correctiosn you must expand the taylor series to the next power:
Answer
given,
Length of pipe, L = 2.65 m
speed of sound, v = 343 m/s
Pipe is open end Pipe
a) Lowest frequency
condition, for open end pipe
For lowest frequency
n = 1
f₁ = 64.72 Hz
b) For second frequency in bugle
n = 2
f₂ = 129.43 Hz
for n = 3
f₃ = 194.16 Hz