Answer:
IUPAC Rules for Alkane Nomenclature
Find and name the longest continuous carbon chain.
Identify and name groups attached to this chain.
Number the chain consecutively, starting at the end nearest a substituent group.
Designate the location of each substituent group by an appropriate number and name.
Explanation:
Answer:
21091mg of aspirin the person need to consume
Explanation:
To solve this question we must find the mass of the person in kg. Knowing the lethal dose for aspirin is 400mg/kg of person, we can find the amount of aspirin that the person need to consume to get a lethal dose:
<em>Mass person:</em>
116lb * (1kg / 2.2lb) = 52.7kg
<em>Lethal dose:</em>
52.7kg * (400mg / kg) =
<h3>21091mg of aspirin the person need to consume</h3>
Eh I can't comment
That's y
No-one knows that I enter this site
Only a cousin knows
Cz I've got a protective family lol
Answer:
10.5 g
Explanation:
Step 1: Given data
- Molar concentration of the solution (C): 0.243 M
- Volume of solution (V): 0.580 L
Step 2: Calculate the moles of solute (n)
Molarity is equal to the moles of solute divided by the liters of solution.
M = n/V
n = M × V
n = 0.243 mol/L × 0.580 L = 0.141 mol
Step 3: Calculate the mass corresponding to 0.141 moles of KCl
The molar mass of KCl is 74.55 g/mol.
0.141 mol × 74.55 g/mol = 10.5 g
Answer:
320 g
Step-by-step explanation:
The half-life of Co-63 (5.3 yr) is the time it takes for half of it to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table as follows:
No. of Fraction Mass
half-lives t/yr Remaining Remaining/g
0 0 1
1 5.3 ½
2 10.6 ¼
3 15.9 ⅛ 40.0
4 21.2 ¹/₁₆
We see that 40.0 g remain after three half-lives.
This is one-eighth of the original mass.
The mass of the original sample was 8 × 40 g = 320 g
