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
GarryVolchara [31]
3 years ago
5

If the caffeine concentration in a particular brand of soda is 3.55 mg/oz, drinking how many cans of soda would be lethal? Assum

e that 10.0 g of caffeine is a lethal dose, and there are 12 oz in a can.
Chemistry
1 answer:
monitta3 years ago
6 0

Answer:

The answer to your question is: 234.7 cans

Explanation:

data

caffeine concentration = 3.55 mg/oz

10.0 g of caffeine is lethal

there are 12 oz of caffeine in a can

Then

                    3.55 mg ----------------- 1 oz

                      x    mg  -----------------12 oz (in a can)

x = 42.6 mg of caffeine in a can

Convert it to grams 42,6 mg = 0.0426 g of caffeine in a can

Finally

            0.0426 g of caffeine ------------------  1 can

            10           g of caffeine -----------------    x

x = 10 x 1/0.0436 = 234.7 cans

You might be interested in
Name two factors that influence the viscosity of a lava flow
mr_godi [17]
Three factors that affect magma viscosity are temperature, composition, and presence of dissolved gases. 
6 0
3 years ago
Give the formula of each coordination compound. Include square brackets around the coordination complex. Do not include the oxid
kogti [31]

Answer:

a) K2[Ni(CN)4]

b) Na3[Ru(NH3)2(CO3)2]

c) Pt(NH3)2Cl2

Explanation:

Coordination compounds are named in accordance with IUPAC nomenclature.

According to this nomenclature, negative ligands end with the suffix ''ato'' while neutral ligands have no special ending.

The ions written outside the coordination sphere are counter ions. Given the names of the coordination compounds as written in the question, their formulas are provided above.

8 0
3 years ago
How many molecules does 1 mole of O2 gas have?
raketka [301]

Answer:

d.6.02×1023 molcules

3 0
3 years ago
Read 2 more answers
How much energy, in joules, does 150.0 g of water with an initial temperature of 25 C need to absorb be raised to a final temper
satela [25.4K]

Answer:

31395 J

Explanation:

Given data:

mass of water = 150 g

Initial temperature = 25 °C

Final temperature = 75 °C

Energy absorbed = ?

Solution:

Formula:

q = m . c . ΔT

we know that specific heat of water is 4.186 J/g.°C

ΔT = final temperature - initial temperature

ΔT = 75 °C - 25 °C

ΔT = 50 °C

now we will put the values in formula

q = m . c . ΔT

q = 150 g × 4.186 J/g.°C × 50 °C

q = 31395 J

so, 150 g of water need to absorb 31395 J of energy to raise the temperature from 25°C to 75 °C .

5 0
3 years ago
Lines on a map joining places that have the same temperature?
vazorg [7]

Answer:

Isotherms

Explanation:

8 0
3 years ago
Read 2 more answers
Other questions:
  • Describe the three temperature scales used in the laboratory and in everyday life: the Fahrenheit, Celsius, and Kelvin scales.
    13·1 answer
  • Which of the following species is a molecular element? Which of the following species is a molecular element? sodium neon carbon
    15·1 answer
  • Heat is transfer directly from one particle of matter to another by the process of
    9·2 answers
  • Describe how protons determine an element's identity .
    10·1 answer
  • Fe2O3(s) + 3CO(
    9·1 answer
  • Given the chemical formula, KNO3, what is the percent nitrogen in the compound?
    10·1 answer
  • When two atoms combine by sharing electrons, it is a(n) _____ bond. <br> ionic<br> covalent
    10·2 answers
  • Can some one please help to understand this problem??<br><br><br>​
    6·1 answer
  • Periodic table the first row of elements fits in the period blank after the element blank the second row of elements fits in per
    13·1 answer
  • What happens if you apply pressure on the oobleck?
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!