Answer: -
(a) 1.54 x 10² pm
(b) 1.3 x 10⁶ is the number of carbon atoms could be aligned side by side in a straight line across the width of 0.2 mm.
Explanation: -
1 å = 10² pm
Diameter of a carbon atom = 1.54 å = 1.54 x 10² pm
1 mm = 10⁹ pm
Total distance = 0.2 mm = 0.2 x 10⁹ pm
Number of carbon atoms could be aligned side by side in a straight line across the width = Total distance / Diameter of a carbon atom
= 
= 1.3 x 10⁶
Answer:
6.73g
Explanation:
T½ = 5.2days
No = 80g
N = ?
T = 20.8days
We'll have to find the disintegration constant first so that we can plug it into the equation that will help us find the mass of the sample after 20.8 days
T½ = In2 / λ
T½ = half life
λ = disintegration constant
λ = In2 / T½
λ = 0.693 / 5.8
λ = 0.119
In(N / No) = -λt
N = final mass of the radioactive sample
No = initial mass of the sample
λ = disintegration constant
t = time for the radioactive decay
In(N/No) = -λt
N / No = e^-λt
N = No(e^-λt)
N = 80 × e^-(0.119 × 20.8)
N = 80 × e^-2.4752
N = 80 × 0.0841
N = 6.728g
The mass of the sample after 20.8 days is approximately 6.73g
Stoichiometry:
First, calculate the number of grams for one mole of Ca3 (PO3)4
(3 * (Mass of Ca)) + (4 * (Mass of P + (3 * Mass of Oxygen)))
= (3*40.08) + 4(30.97 + (3*16.00))
=(120.24) + 4(78.97)
=436.12 g / mol Ca3(PO3)4
This means there are 436.12 g per 1 mole of Ca(PO3)4. Since there are 4.50 moles of Calcium Phosphate, mulitply the molar mass of Ca(PO3)4 by 4.50 and you should get 1962.54 g. Since there are 3 sigfigs, the final answer is 1960 g.
on a side note: I put in all my work in case 1. your periodic table if different, 2. my work is wrong, 3. you put in the question wrong because I feel that the actual compound would be Ca3(PO4)3 instead of Ca3(PO3)4 (if this is the case, the answer should be 1820 g).
Molar mass of CeCl3 = 246.475 g/mol
This compound is also known as Cerium(III) Chloride.
Convert grams CeCl3 to moles or moles CeCl3 to grams
Molecular weight calculation:
140.116 + 35.453*3
