Answer:
Iron remains = 17.49 mg
Explanation:
Half life of iron -55 = 2.737 years (Source)
Where, k is rate constant
So,
The rate constant, k = 0.2533 year⁻¹
Time = 2.41 years
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
= 32.2 mg
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
is the initial concentration
So,
![[A_t]=32.2\times e^{-0.2533\times 2.41}\ mg](https://tex.z-dn.net/?f=%5BA_t%5D%3D32.2%5Ctimes%20e%5E%7B-0.2533%5Ctimes%202.41%7D%5C%20mg)
![[A_t]=32.2\times e^{-0.610453}\ mg](https://tex.z-dn.net/?f=%5BA_t%5D%3D32.2%5Ctimes%20e%5E%7B-0.610453%7D%5C%20mg)
![[A_t]=17.49\ mg](https://tex.z-dn.net/?f=%5BA_t%5D%3D17.49%5C%20mg)
<u>Iron remains = 17.49 mg</u>
6.022×10^23 should be correct. Are there any options to choose from?
<u>Avogadros number</u>
Answer:
Mass of water = 73.08 g
Explanation:
Given data:
Mass of hydrogen = 35 g
Mass of oxygen = 65 g
Mass of water = ?
Solution:
First of all we will write the balanced chemical equation:
2H₂ + O₂ → 2H₂O
Number of moles of hydrogen = mass/ molar mass
Number of moles of hydrogen = 35 g/ 2 g/mol
Number of moles of hydrogen = 17.5 mol
Number of moles of oxygen = 65 g / 32 g/mol
Number of moles of oxygen = 2.03 moles
Now we compare the moles of water with moles hydrogen and oxygen.
H₂ : H₂O
2 : 2
17.5 : 17.5
O₂ : H₂O
1 : 2
2.03 : 2× 2.03 =4.06 mol
Number of moles of water produced by oxygen are less so oxygen is limitting reactant.
Mass of water:
Mass of water = number of moles × molar mass
Mass of water = 4.06 mol × 18 g/mol
Mass of water = 73.08 g
