The correct answer is (3)
I-131 and P-32
The explanation:
according to attached table:
- we can see that the half life of p 32 is 14.28d (more than one hour)
- and the half life of I-131 is 8.021 d
(more than one hour)
and They both have β- decay mode and with half-lives greater than hour.
<h3>
Answer:</h3>
16.7 g H₂O
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Stoichiometry</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 2NaOH (s) + CO₂ (g) → Na₂CO₃ (s) + H₂O (l)
[Given] 1.85 mol NaOH
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol NaOH → 1 mol H₂O
Molar Mass of H - 1.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol
<u>Step 3: Stoichiometry</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
16.6685 g H₂O ≈ 16.7 g H₂O
Answer:
30 cm³
Explanation:
Step 1: Given data
- Density of aluminum (ρ): 2.7 g/cm³
- Mass of aluminum (m): 81 g
- Volume occupied by aluminum (V): ?
Step 2: Calculate the volume occupied by aluminum
The density of aluminum is equal to its mass divided by its volume.
ρ = m/V
V = m/ρ
V = 81 g / 2.7 g/cm³
V = 30 cm³
The production of new skin cells is example of regeneration.
Regeneration is process of replacing or restoring not just skin cells, but also other cells in the human organism.
Skin regeneration is replacement of damaged tissue with new tissue.
There are two ways of skin regeneration:
1) reconstruction is a process of rebuilding of damaged skin cells
2) restoration is process of replacing broken cell skins
Epidermal stem cells are those who produces new daughter cell skins.
Epidermal stem cells are in the lowest layer of the skin.
More about skin: brainly.com/question/306377
#SPJ4
Answer : 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.
Solution : Given,
Mass of Cu = 300 g
Molar mass of Cu = 63.546 g/mole
Molar mass of 
= 183.511 g/mole
- First we have to calculate the moles of Cu.
The moles of Cu = 4.7209 moles
From the given chemical formula, we conclude that the each mole of compound contain one mole of Cu.
So, The moles of Cu = Moles of = 4.4209 moles
- Now we have to calculate the mass of .
Mass of = Moles of × Molar mass of = 4.4209 moles × 183.511 g/mole = 866.337 g
Mass of = 866.337 g = 0.8663 Kg (1 Kg = 1000 g)
Therefore, 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.
