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kirill [66]
3 years ago
12

Suppose that you take 200 mg of an antibiotic every 8 hr. The​ half-life of the drug is 8 hr​ (the time it takes for half of the

drug to be eliminated from your​ blood). Use infinite series to find the​ long-term (steady-state) amount of antibiotic in your blood exactly.
Chemistry
1 answer:
DochEvi [55]3 years ago
5 0

Answer:

600 mg

Explanation:

The initial amount of the drug = 200 mg

The half-life of the drug = 8 hrs

It means that:-

After 6 hours, the concentration becomes :- \frac{200}{2} mg

After 12 hours, the concentration becomes :- \frac{200}{4} mg

After 18 hours, the concentration becomes :- \frac{200}{8} mg

And so on...

Thus,

After infinite time = 200+\frac{200}{2}+\frac{200}{4}+\frac{200}{8}+..

Thus,

After infinite time = 200\times (1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+..)

The sum of the infinite series is:- 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.. = \frac{1}{1+\frac{1}{2}}=2

So,

<u>After infinite time = 200\times 2 mg = 600 mg</u>

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In a certain experiment, 28.0 mL of 0.250 M HNO3and 53.0 mL of 0.320 M KOH are mixed. Calculate the number of moles of water for
Xelga [282]

Answer:

The number of moles of water formed in the resulting reaction is 6.03

[H+]: 37,2 M

[OH-]: 37,2 M

Explanation:

HNO3  +  KOH ----> KNO3 + H2O

First, we must discover the limiting reagent and we need to find out the moles, we use for this.

Moles that are used = Molarity / volume

HNO3 : 0,250 mol/L / 0,028L = 8,93 moles

KOH : 0,320 mol/L / 0,053L = 6,03 moles

The ratio of the reagents by stoichiometry is 1 to 1, so the limiting reagent is KOH, if I need 1 mole of nitric per mole of KOH, for every 8.93 moles I will need the same. However I have only 6.03 moles of KOH

The ratio of the reagents/products by stoichiometry is 1 to 1 so if I need 1 mol of KOH to make 1 mol of Water, 6,03 moles of KOH are used to make 6,03 moles of H2O.

The equilibrium of water is this:

2H2O ⇄ H3O+  +  OH-

2 moles of water are broken down into 1 mole of hydronium (H3O +) and 1 mole of hydroxyl (OH-)

6,03 moles of water are broken down into the half of those moles, so we have 3,015 moles of H3O+ and 3,015 moles of OH- but these moles are in 81,0 mL (the volume of the two solutions, 28 mL + 53 mL)

We must find out the moles in 1000 mL (1 L) so let's apply the rule of three.

81 mL ____ 3,015 moles

1000 mL ___ ( 1000 . 3,015) /81 = 37,2 M

7 0
2 years ago
HELPPP!!! In reaction 2Sr+O2=2SrO how many moles of Oxygen will be needed to react with 36.9g of SrO?
Step2247 [10]

6000 I pretty sure tell me if it's right

6 0
2 years ago
I need part 1 and 2 please , just separate answers
Vladimir79 [104]

First, we have to remember the molarity formula:

M=\text{ }\frac{moles\text{ of solute}}{L\text{ solution}}

Part 1:

In this case, our solute is sodium nitrate (NaNO3), and we have the mass dissolved in water, then we have to convert grams to moles. For that, we need the molecular weight:

M.W_{NaNO_3}=\text{ 23+14+16*3= 85 g/mol}

Then, we calculate the moles present in the solution:

3.976\text{ g NaNO}_3\text{ * }\frac{1\text{ mol}}{85\text{ g}}=\text{ 0.04678 mol NaNO}_3

Now, we have the necessary data to calculate the molarity (with the solution volume of 200 mL):

M=\frac{0.04678\text{ mol}}{200\text{ mL*}\frac{1\text{ L}}{1000\text{ mL}}}=\text{ 0.2339 M}

The molarity of this solution equals 0.2339 M.

Part 2:

In this case, we have the same amount (in moles and mass) of sodium nitrate, but a different volume of solution, then we only have to change it:

M=\text{ }\frac{0.04678\text{ mol}}{275\text{ mL *}\frac{1\text{ L}}{1000\text{ mL}}}=\text{ 0.1701 M}

So, the molarity of this solution is 0.1701 M.

5 0
11 months ago
For a phase change, H0 = 2 kJ/mol and A S0 = 0.017 kJ/(K•mol). What are
34kurt

Answer:

ΔG = -6.5kJ/mol at 500K

Explanation:

We can find ΔG of a reaction using ΔH, ΔS and absolute temperature with the equation:

ΔG = ΔH - TΔS

Computing the values in the problem:

ΔG = ?

ΔH = 2kJ/mol

T = 500K

And ΔS = 0.017kJ/(K•mol)

Replacing:

ΔG = 2kJ/mol - 500K*0.017kJ/(K•mol)

ΔG = 2kJ/mol - 8.5kJ/mol

<h3>ΔG = -6.5kJ/mol at 500K</h3>

8 0
2 years ago
Determine the formula weight of BeCl₂.
tia_tia [17]
Be= Beryllium= weight 9.01
Cl= Chlorine= weight 35.45
Since there are 2 Chlorine, you have to add Chlorine twice.
9.01 + 35.45 + 35.45= 79.91

Answer: 79.91
(This answer does not include sig.figs)
4 0
3 years ago
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