1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Ad libitum [116K]
3 years ago
8

A certain drug has a half-life in the body of 4.0h. What should the interval between doses be, if the concentration of drug in t

he body should not fall below 15.% of its initial concentration? Round your answer to 2 significant digits.
Chemistry
1 answer:
scZoUnD [109]3 years ago
8 0

Answer:

Explanation:

half life = 4 h

initial concentration = 100

final concentration = 15

Time = t

No of half life  x = t / 4

15 = 100 x ( 1/2 )ˣ

.15 =  ( 1/2 )ˣ

ln .15 = - x ln2

x = - ln .15 / ln 2

= 1.897 / .693

x = 2.737

x = t / 4

t = 2.737 x 4 = 11 h approx .

You might be interested in
How many moles are in 6.80 x 10^23 atoms <br> of gold, Au?
frosja888 [35]

Answer:

1.13 moles Au

Explanation:

Moles Au = 6.80x10²³atoms / 6.023x10²³atoms/mole = 1.13 moles Au

8 0
2 years ago
What are 2 diseases caused by coronaviruses?
ivann1987 [24]

Answer:

SARS and MERS

Explanation:

8 0
3 years ago
Read 2 more answers
Calculate the hydroxide ion concentration [OH-] for a solution with a pH of 6.10
nalin [4]

There are different formula you need to keep in mind when solving for [OH-]

Given that pH = 6.10

pH + pOH = 14

6.10 + pOH = 14

pOH = 7.9

[OH-] = 10^(-pOH)

[OH-] = 10^(-7.9)

[OH-] = 0.000000013

[OH-] = 1.3 x 10^-8


<h2><u>Answer: [OH-] = 1.3 x 10^-8</u></h2>
4 0
2 years ago
What is the molarity of a NaOH solution if 32.47 mL is required to titrate 0.6013 g of potassium hydrogen phthalate (C8H5O4K)
Artist 52 [7]

Answer:

0.0907 M

Explanation:

Before you can calculate the molarity, you need to convert grams to moles (via molar mass) and convert mL to L.

(Step 1)

Molar Mass (C₈H₅O₄K):

8(12.011 g/mol) + 5(1.008 g/mol) + 4(15.998 g/mol) + 39.098 g/mol

Molar Mass (C₈H₅O₄K): 204.218 g/mol

0.6013 g C₈H₅O₄K             1 mole
------------------------------  x  ------------------  =  0.00294 moles C₈H₅O₄K
                                          204.218 g

(Step 2)

1,000 mL = 1 L

32.47 mL              1 L
---------------  x  -----------------  =  0.03247 L
                        1,000 mL

(Step 3)

Molarity (M) = moles / volume (L)

Molarity = 0.00294 moles / 0.03247 L

Molarity = 0.0907 M

6 0
1 year ago
When a sodium (Na) atom forms an ion so that it looks like a noble gas with a full valence shell, what will its charge be?
Setler79 [48]

Answer:

I believe 1+

Explanation:

when Na loses 1 electron in the outer shell it has 8 valence protons on it's new most outer shell. so now it has 11 protons and 10 electrons. that extra proton (positively charged) adds one extra charge. so +1

5 0
3 years ago
Other questions:
  • Salinity is a measure of which of the following in water?
    6·2 answers
  • List two things that do not change between isotopes of an element
    14·2 answers
  • Consider the formation of hcn by the reaction of nacn (sodium cyanide) with an acid such as h2so4 (sulfuric acid): 2nacn(s)+h2so
    10·2 answers
  • 5. Astronomers recently discovered a moon that does not<br> orbit a planet.
    13·1 answer
  • Which process is an example of a chemical change? drying clothes in the dryer heating a cup of tea chopping a tree activating a
    8·1 answer
  • Is corona increasing or decreasing the albedo effect? Explain.
    13·2 answers
  • Which of the following is a common property of both strong acid and strong base?
    6·2 answers
  • According to the enthalpy diagram below, which of the following statements is true?
    5·2 answers
  • What is the main purpose of the concentration gradient and why is it vital to the process of osmosis?
    14·2 answers
  • Why doesn’t the addition of an acid-base indicator affect the pH of the test solution?
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!