To solve this problem we will apply the concept of wavelength, which warns that this is equivalent to the relationship between the speed of the air (in this case in through the air) and the frequency of that wave. The air is in standard conditions so we have the relation,
Frequency 
Speed of sound in air 
The definition of wavelength is,
Here,
v = Velocity
f = Frequency
Replacing,
Therefore the wavelength of that tone in air at standard conditions is 0.589m
Answer:
The answer is the option a.
Explanation:
We know that magnetic force (Fm) is defined as
Fm = q (v x B)
Where q is a the value of the charge, v is the velocity of the charge and B is the value of the magnetic field.
"v x B" is defined as the cross product between the vectors velocity and magnetic field, and if the angle between them is thetha < 180°, then, the cross product is
v x B = vBsin (thetha)
So,
Fm = qvBsin (thetha)
And, in case in which v and B are parallel vectors, thetha is zero, and,
sin (thetha)=sin (0) = 0
So, Fm=0
The environmental changes can will negatively affect different organisms differently, for example some organisms might die due to failure of the organisms to adapt to the new changes. Some organisms will be forced to migrate in order to counter these changes, this may lead to this organisms leaving their old living routine and adapt to the new methods of living, this may lead to evolution caused by new environment.
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.
