Answer:
Due presence of empty as well as heavy particles.
Explanation:
The main reason for this is the presence of empty spaces as well as heavy particles in the nucleus of Helium atom. Most of the beam pass through the helium atom due to the presence of empty spaces in the atom whereas some beams deflect from the helium atom because both repel each other due to same charge i.e. positive charge. Few beams of light bounce back with an angle of 90 degree due hitting with the nucleus where heavy particles such as neutron and proton are present.
D the periodic table shows the elements and their atoms and how they are arranged
The balance chemical equation is as follow,
2 Al + 3 O₂ → Al₂O₃
Aluminium is the Limiting Reagent,
As,
107.92 g Al required = 96 g of O₂
Then,
82.49 g of Al will require = X g of O₂
Solving for X,
X = (82.49 g × 96 g) ÷ 107.92 g
X = 73.37 g of O₂
But,
We are provided with 117.65 g of O₂, So, it is provided in excess and 44.28 g of it will remain unreacted.
Solving for Amount of Al₂O₃ formed,
As,
107.92 g of Al produced = 203.92 g of Al₂O₃
Then,
82.49 g of Al will produce = X g of Al₂O₃
SOlving for X,
X = (82.49 g × 203.92 g) ÷ 107.92
X = 155.86 g of Al₂O₃
Answer:
The answer to your question is 0.34 M
Explanation:
Data
[Sr(OH)₂] = ?
Volume of Sr(OH)₂ = 35.6 ml or 0.0356 l
[HBr] = 0.0445 M
Volume of HBr = 0.549 l
Balanced chemical reaction
2HBr + Sr(OH)₂ ⇒ SrBr₂ + 2H₂O
Process
1.- Calculate the moles of HBr
Molarity = moles / volume
-Solve for moles
moles = Molarity x volume
-Substitution
moles = 0.0445 x 0.549
= 0.0244
2.- Calculate the moles of Sr(OH)₂ using the coefficients of the balanced reaction
2 moles of HBr ----------------------- 1 mol of Sr(OH)₂
0.0244 moles ----------------------- x
x = (0.0244 x 1) / 2
x = 0.0122 moles of Sr(OH)₂
3.- Calculate the concentration of Sr(OH)₂
Molarity = 0.0122/ 0.0356
-Simplification
Molarity = 0.34
