Answer:
1 mole of Al2O3 = 102 grams
1 mole of Al2 = 54 grams
102 grams of Al2O3 contains = 54 gram of Al2
10kg of Al2O3 contains = (54/102)*10000g Al2
= 5294.11 g Al2 or 5.29411 kg
<span>Of the answers listed option B looks like the most complete. Ie "Check for the presence of alpha, beta, and gamma particles." the significant presence of these particles is a specific indicator of radioactive decay, i.e: unstable atoms spontaneously undergoing a nuclear reaction.</span>
Answer:
9.25
Explanation:
Let first find the moles of 
and 
number of moles of = 0.40 mol/L × 200 × 10⁻³L
= 0.08 mole
number of moles of = 0.80 mol/L × 50 × 10⁻³L
= 0.04 mole
The equation for the reaction is expressed as:
The ICE Table is shown below as follows:
Initial (M) 0.08 0.04 0
Change (M) - 0.04 -0.04 + 0.04
Equilibrium (M) 0.04 0 0.04
for buffer solutions
since they are in the same solution
BBBBBBBBBBBBBBBBBBBBBBBBBBBBB
Answer:
Q = 3440Kj
Explanation:
Given data:
Mass of gold = 2kg
Latent heat of vaporization = 1720 Kj/Kg
Energy required to vaporize 2kg gold = ?
Solution:
Equation
Q= mLvap
It is given that heat required to vaporize the one kilogram gold is 1720 Kj thus, for 2 kg
by putting values,
Q= 2kg × 1720 Kj/Kg
Q = 3440Kj
