425 g X 1 mL/.92 g = 462 mL
Answer:
The amount of energy transferred to the diamond while being cut is thus Q = 852000 J
Explanation:
Since quantity of heat transferred Q = mcΔθ where m = mass of substance , c = specific heat capacity of substance and Δθ = temperature change.
Now, given that for diamond, m = mass of diamond = 600 g, c = specific heat capacity of diamond = 710 J/g°C and Δθ = temperature change = 2 °C.
So, the amount of energy transferred to the diamond while being cut is thus
Q = mcΔθ
Q = 600 g × 710 J/g°C × 2 °C
Q = 852000 J
So, the amount of energy transferred to the diamond while being cut is thus Q = 852000 J
Answer:
There were originally 8 atoms of Potassium-40.
Explanation:
The half-life of a radioactive material is the time taken for half the original material to decay or the time required for a quantity of the radioactive substance to reduce to half of its initial value.
If the original material formed without any Argon-40, it means that the atoms originally present were Potassium-40 atoms.
Presently, there are 7 Argon-40 atoms for every 1 of Potassium-40, we can deduce the number of half-lifes the Potassium-40 has undergone as follows :
After one half-life, (1/2) there will be one Potassium-40 atom for every Argon-40 atom.
After a second half life, 1/2 × 1/2 = 1/4: there will be one Potassium-40 atom for every three atoms of Argon-40.
After a third half-life, 1/4 × 1/2 = 1/8: there will be one Potassium-40 atom for every 7 atoms of Argon-40.
Since there are 1/8 atoms of Potassium-40 presently, there were originally 8 atoms of Potassium-40.
Explanation:
such bad questions , how it could be around ???
Answer:
71.372 g or 0.7 moles
Explanation:
We are given;
- Moles of Aluminium is 1.40 mol
- Moles of Oxygen 1.35 mol
We are required to determine the theoretical yield of Aluminium oxide
The equation for the reaction between Aluminium and Oxygen is given by;
4Al(s) + 3O₂(g) → 2Al₂O₃(s)
From the equation 4 moles Al reacts with 3 moles of oxygen to yield 2 moles of Aluminium oxide.
Therefore;
1.4 moles of Al will require 1.05 moles (1.4 × 3/4) of oxygen
1.35 moles of Oxygen will require 1.8 moles (1.35 × 4/3) of Aluminium
Therefore, Aluminium is the rate limiting reagent in the reaction while Oxygen is the excess reactant.
4 moles of aluminium reacts to generate 2 moles aluminium oxide.
Therefore;
Mole ratio Al : Al₂O₃ is 4 : 2
Thus;
Moles of Al₂O₃ = Moles of Al × 0.5
= 1.4 moles × 0.5
= 0.7 moles
But; 1 mole of Al₂O₃ = 101.96 g/mol
Thus;
Theoretical mass of Al₂O₃ = 0.7 moles × 101.96 g/mol
= 71.372 g