Answer:
a new substance is formed, if heat is formed, color change, the evolution of gas, and bubbles or fizzing
Pressure can be defined as the force acting on a perpendicular surface per unit area.
Force exerted by a man of mass 100 kg wearing snow shoes = m.a
Where m = mass of the man = 100 kg
a = acceleration due to gravity= 9.8 
Force exerted by the man of mass 100 kg = 
Force exerted by woman of mass 60 kg = 
Force exerted by 100 kg man is greater than that exerted as 60 kg woman. The area on which this force is acting determines the pressure. Pressure is inversely proportional to the area on which the force acts. Therefore, the pressure exerted by 100 kg man wearing snow shoes is less than the pressure exerted by a 60 kg woman woman wearing high heels as the force acts over a larger area when the man wears snow shoes when compared to the force exerted over a smaller area in case of the woman wearing high heels.
The reaction for what was describe in the problem is:
N₂ + 3 O₂ --> 2 NO₃
The reactants involved are nitrogen and oxygen gas. From the word itself, oxygen is an oxidizing agent. <em>Therefore, this reaction is an oxidation reaction due to the presence of the oxidizing agent.</em>
Answer:
Explanation:
Given that
d= 35 μm ,yield strength = 163 MPa
d= 17 μm ,yield strength = 192 MPa
As we know that relationship between diameter and yield strength


d = diameter
K =Constant

So now by putting the values
d= 35 μm ,yield strength = 163 MPa
------------1
d= 17 μm ,yield strength = 192 MPa
------------2
From equation 1 and 2

K=394.53
By putting the values of K in equation 1


Now when d= 12 μm


Answer:
The volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L
Explanation:
Given data:
Number of moles of HF = 6.62×10⁻³ mol
Volume of HF in litter at STP = ?
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Standard temperature = 273 K
Standard pressure = 1 atm
Now we will put the values in formula.
1 atm × V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K × 273 K
V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K × 273 K / 1 atm
V = 148.38×10⁻³ L
Thus, the volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L.