Depends on the thoroughness of that particular experiment. But just in general no, you usually need more than one experiment to verify a hypothesis.
Answer:
The mass of KClO₃ that will absorb the same heat as 5 g of KCl is 3.424 g
Explanation:
Here we have
Heat of solution of KClO₃ = + 41.38 kJ/mol.
Heat of solution of KCl (+17.24 kJ/mol)
Therefore, 1 mole of KCl absorbs +17.24 kJ during dissolution
Molar mass of KCl = 74.5513 g/mol
Molar mass of KClO₃ = 122.55 g/mol
74.5513 g of KCl absorbs +17.24 kJ during dissolution, therefore, 5 g will absorb

Therefore the amount of KClO₃ to be dissolved to absorb 1.156 kJ of energy is given by
122.55 g of KClO₃ absorbs + 41.38 kJ, therefore,

Therefore the mass of KClO₃ that will absorb the same heat as 5 g of KCl = 3.424 g.
2Al + 3Fe(NO₃)₂ = 3Fe + 2Al(NO₃)₃
m=245 g
w=0.805 (80.5%)
M{Fe(NO₃)₂}=179.857 g/mol
M(Fe)=55.847 g/mol
1. the mass of salt in solution is:
m{Fe(NO₃)₂}=mw
2. the proportion follows from the equation of reaction:
m(Fe)/3M(Fe)=m{Fe(NO₃)₂}/3M{Fe(NO₃)₂}
m(Fe)=M(Fe)m{Fe(NO₃)₂}/M{Fe(NO₃)₂}
m(Fe)=M(Fe)mw/M{Fe(NO₃)₂}
m(Fe)=55.847*245*0.805/179.857= 61.24 g
The density of a material is its unique property by which a unknown material can be identified and also the impurity (if present) in a material can be concluded. Mathematically the density can be expressed as-
. Thus from the mathematical expression we can say that the density is the mass per unit volume of a material. Here the density of ethanol is given 0.789 g/mL. Thus the weight of the 1 mL ethanol is 0.789g. Thus the weight of the 125 mL of ethanol will be (125×0.789) = 98.625 g.
Answer:


Explanation:
From the question we are told that:
Edge length of the unit cell 
a)
Generally the equation for The relationship between edge length and radius is mathematically given by

Therefore



b)
From the question we are told that:
Density 
Edge length of 
Therefore Volume is given as



Generally the equation for Mass is mathematically given by




Therefore Molarity is given as



Finally The atoms in a unit cell is


