The carbon tetrachloride molecule CCI 4 is

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

DochEvi [55]

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,

After infinite time = $200\times 2$ mg = 600 mg

5 0

3 years ago

Non-renewable energy is often called?

DIA [1.3K]

Answer:

It is also called a finite resource

Explanation:

5 0

3 years ago

Someone pls help me I will make you brain

Olegator [25]

Answer:

the second answer

Explanation:

5 0

3 years ago

Read 2 more answers

How many significant figures are in this number? 0.00708900

Norma-Jean [14]

Don’t know sorry 1.627

4 0

3 years ago

Read 2 more answers

Show All Work For Brainliest

sattari [20]

Answer:

1) 16.8 L CO2

2) 36.96 L NH3

3) 9.88 L CO2

4) 56.99 L H2O

Explanation:

How many liters of carbon dioxide gas will be produced when 75.0 g of calcium carbonate decomposes to form calcium oxide when at STP?

CaCO3 → CaO + CO2

Moles calcium carbonate = 75.0 grams / 100.09 g/mol

Moles calcium carbonate = 0.750 moles

For 1 mol CaCO3 we'll have 1 mol CaO and 1 mol CO2

For 0.750 moles CaCO3 we'll have 0.750 moles CO2

1 mol = 22.4 L

0.750 moles CO2 = 0.750 *22.4 L = 16.8 L CO2

2. Hydrogen gas reacts with 23.1 g of nitrogen gas to produce ammonia (NH3). What volume of ammonia will be produced at STP?

3H2 + N2 → 2NH3

Moles N2 = 23.1 grams / 28.0 g/mol

Moles N2 = 0.825 moles

For 3 moles H2 we need 1 mol N2 to produce 2 moles NH3

For 0.825 moles N2 we'll have 2*0.825 = 1.65 moles NH3

1 mol = 22.4 L

1.65 mol = 1.65 * 22.4 L = 36.96 L NH3

3. Iron (III) oxide reacts with carbon monoxide to form iron and carbon dioxide. How many liters of carbon dioxide will be produced from 23.5 g of iron (III) oxide when at STP?

Fe2O3 + 3CO → 2Fe + 3CO2

Moles Fe2O3 = 23.5 grams / 159.69 g/mol

Moles Fe2O3 = 0.147 moles

For 1 mol Fe2O3 we need 3 moles CO to produce 2 moles Fe and 3 moles CO2

For 0.147 moles Fe2O3 we'll have 3*0.147 = 0.441 moles CO2

1 mol = 22.4 L

0.441 moles = 22.4 * 0.441 = 9.88 L CO2

4.How many liters of water vapor would be produced in the combustion of 12.5L of ethane, C2H6 at STP?

2C2H6 + 7O2 →4CO2 + 6H2O

22.4 L = 1 mol

12.5 L = 0.848 moles C2H6

For 2 moles C2H6 we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

For 0.848 moles C2H6 we'll have 3*0.848 = 2.544 moles H2O

1 mol = 22.4 L

2.544 moles = 22.4 L * 2.544 = 56.99 L H2O

5 0

3 years ago