1. to calculate molar mass , you need average atomic mass of each element.
formula is simple
suppose we have Na2CO3
here it says
there are
2 atoms of Na
1 atom of C
3 atom of O
now
molar mass = (total atoms of Na × average atomic mass of Na) + ( total atom of C × average atomic mass of C )+ (total atom of O × average atomic mass of O)
i hope you can figure out it now.
in case of (NH4)2
this means you have 2 count of NH4
so total atoms of N = 2
total atoms of H = 4×2 = 8
so on you can now figure it out
2. balanced equation
here LHS and RHS shoild have same number of atoms of each element .
for example
H2 + Cl2 => 2HCl
here 2 atoms of H combined with 2 atoms of Cl , we know H and Cl combined and formed HCl now because total H atoms are 2 and Cl atoms are 2 in LHS then we should also have 2 atoms of each in right side
so thats why we have 2HCl in right side.
Try to figure out . let me know if need any help
Answer:
6.95 x 10²³ molecules/particles
Explanation:
First we need to find the total Empirical Mass. We can find this by adding each element's mass together.
Al = 26.982,
O = 15.999
H = 1.008
26.982 + 3(15.999) + 3(1.008) = 78.003.
Now we divide by the mass given (90 grams).
90/78.003 = 1.153801777.
We then take that number and multiply it by avogadro's number (6.022 x 10²³)
1.153801777 x avogadro's number = 6.95 x 10²³
C. Tripling the length and reducing the radius by a factor of 2 is the change to a pipe would increase the conductance by a factor of 12.
<u>Explanation:</u>
As we know that the resistance is directly proportional to the length of the pipe and it is inversely proportional to the cross sectional area of the pipe.
So it is represented as,
R∝ l /A [ area is radius square]
So k is the proportionality constant used.
R = kl/A
Conductance is the inverse of resistance, so it is given as,
C= 1/R.
R₁ = kl₁ / A₁
R₂ = kl₂/A₂
R₂/R₁ = 1/12 [∵ conductance is the inverse of resistance]
= l₂A₁ / l₁A₂
If we chose l₁/l₂= 3 and A₂/A₁= 4 So R₂/R₁= 1/3×4 = 1/12
So tripling the length and reducing the radius by a factor of 2 would increase the conductance by a factor of 12.
