Answer:
The amount of drug left in his body at 7:00 pm is 315.7 mg.
Explanation:
First, we need to find the amount of drug in the body at 90 min by using the exponential decay equation:
Where:
λ: is the decay constant = 
: is the half-life of the drug = 3.5 h
N(t): is the quantity of the drug at time t
N₀: is the initial quantity
After 90 min and before he takes the other 200 mg pill, we have:
Now, at 7:00 pm we have:
Therefore, the amount of drug left in his body at 7:00 pm is 315.7 mg (from an initial amount of 400 mg).
I hope it helps you!
<em>Answer:</em>
- 0.052301 km have 5 significant figure
- 400 cm have 1 significant figure
- 50.0 m have 3 significant figure
- 4500.01 ml have 6 significant figure
<em>Explanation:</em>
According to rules of significant figure
0.052301 km have 5 significant figure:
- Zero to the left of the first non zero digit not significant.
- Zero between the non zero digits are significant.
<em>400 cm have 1 significant figure:</em>
- Trailing zeros are not significant in numbers without decimal points.
<em>50.0 m have 3 significant figure:</em>
- Trailing zeros are significant in numbers when there is decimal points.
<em>4500.01 ml have 6 significant figure:</em>
- Zero between the non zero digits are significant.
Answer:
If there reacted 1.5 moles of O2, there will be produced 1.0 mol of Fe2O3
Explanation:
Step 1: Data given
Number of moles oxygen reacted = 1.5 moles
Step 2: The balanced equation
4Fe + 3O2 → 2Fe2O3
Step 3: Calculate moles of Fe2O3
For 4 moles Fe consumed, we need 3 moles of O2 to produce 2 moles of Fe2O3
For 1.5 moles O2 consumed, we'll have 2/3 * 1.5 = 1.0 mol of Fe2O3
If there reacted 1.5 moles of O2, there will be produced 1.0 mol of Fe2O3
25 gigaseconds is equal to 2,500,000,000,000 centiseconds
Calcium bicarbonate - Ca(HCO3)2
sodium peroxide - Na202
water - H20
silver nitrate - AgN03
potassium carbonate - K2CO3
sodium carbonate - Na2CO3
zinc chloride - ZnC12
calcium hydroxide - Ca(OH)2
magnesium chloride - MgC12
