First a balanced reaction equation must be established:
→
Now if mass of aluminum = 145 g
the moles of aluminum = (MASS) ÷ (MOLAR MASS) = 145 g ÷ 30 g/mol
= 4.83 mols
Now the mole ratio of Al : O₂ based on the equation is 4 : 3
[
4Al +
3 O₂ → 2 Al₂O₃]
∴ if moles of Al = 4.83 moles
then moles of O₂ = (4.83 mol ÷ 4) × 3
=
3.63 mol (to 2 sig. fig.)
Thus it can be concluded that
3.63 moles of oxygen is needed to react completely with 145 g of aluminum.
Answer:
The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)
Explanation:
Step 1: Data given
Initial temperature = 10.0 °C
Final temperature = 25.0 °C
Energy required = 30000 J
Mass of the object = 40.0 grams
Step 2: Calculate the specific heat capacity of the object
Q = m* c * ΔT
⇒With Q = the heat required = 30000 J
⇒with m = the mass of the object = 40.0 grams
⇒with c = the specific heat capacity of the object = TO BE DETERMINED
⇒with ΔT = The change in temperature = T2 - T2 = 25.0 °C - 10.0°C = 15.0 °C
30000 J = 40.0 g * c * 15.0 °C
c = 30000 J / (40.0 g * 15.0 °C)
c = 50 J/g°C
The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)
Answer :
121.5 <span>
μCi
Explanation : We have Ce-141 half life given as 32.5 days so if the activity is 3.8 </span><span>μci after 162.5 days of time elapsed we have to find the initial activity.
We can use this formula;
</span>
3.8 /
=
((0.693 X 162.5 ) / 32.5) = 121.5
<span>
On solving we get, The initial activity as 121.5 </span>μci
Answer : The given compound belongs to ether and alcohol.
Explanation :
The chemical formula of the given compound is, 
First we have to calculate the degree of unsaturation.
Formula used:
Degree of unsaturation = 
where,
C = number of carbon
H = number of hydrogen
N = number of nitrogen
X = number of halogen
Degree of unsaturation = 
The degree of unsaturation is, 0 that means there is no double or triple bond in the compound only single bond is present between the atoms.
Thus, the given compound belongs to ether and alcohol.
Answer: 122 moles
Procedure:
1) Convert all the units to the same unit
2) mass of a penny = 2.50 g
3) mass of the Moon = 7.35 * 10^22 kg (I had to arrage your numbers because it was wrong).
=> 7.35 * 10^22 kg * 1000 g / kg = 7.35 * 10^ 25 g.
4) find how many times the mass of a penny is contained in the mass of the Moon.
You have to divide the mass of the Moon by the mass of a penny
7.35 * 10^ 25 g / 2.50 g = 2.94 * 10^25 pennies
That means that 2.94 * 10^ 25 pennies have the mass of the Moon, which you can check by mulitiplying the mass of one penny times the number ob pennies: 2.50 g * 2.94 * 10^25 = 7.35 * 10^25.
5) Convert the number of pennies into mole unit. That is using Avogadros's number: 6.022 * 10^ 23
7.35 * 10^ 25 penny * 1 mol / (6.022 * 10^ 23 penny) = 1.22* 10^ 2 mole = 122 mol.
Answer: 122 mol
