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zimovet [89]
3 years ago
9

You are preparing to work with Chemical A. You open the appropriate storage cabinet, and notice Chemical B, as well as Chemical

A. Based on your knowledge of both chemicals you suspect that they should not be stored together. Which section of the SDS should you consult, in order to confirm your suspicion?
Engineering
1 answer:
PtichkaEL [24]3 years ago
5 0

The correct answer is; Stability and reactivity.

Further Explanation:

The stability and reactivity section of the SDS sheets is where to check for the possibility of hazardous reactions for the chemicals. This also lists the chemical stability of each chemical that people may be using. This can be found in section 10 of the OSHA Quick Card.

The SDS sheets has 16 sections for employees to use. Since 2015, the sections can be found in uniform format for easier and faster ways to find the section needed. The 16 sections for the SDS sheets are:

  1. Identification
  2. Hazard(s) identification
  3. Composition/information on ingredients
  4. First-aid measures
  5. First-aid measures
  6. First-aid measures
  7. Handling and storage
  8. Exposure controls/personal protection
  9. Physical and chemical properties
  10. Stability and reactivity
  11. Toxicological information
  12. Ecological information
  13. Disposal considerations
  14. Transport information
  15. Regulatory information
  16. Other information

Learn more about SDS sheets at brainly.com/question/9753408

#LearnwithBrainly

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Small droplets of carbon tetrachloride at 68 °F are formed with a spray nozzle. If the average diameter of the droplets is 200 u
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Answer:

the difference in pressure between the inside and outside of the droplets is 538 Pa

Explanation:

given data

temperature = 68 °F

average diameter = 200 µm

to find out

what is the difference in pressure between the inside and outside of the droplets

solution

we know here surface tension of carbon tetra chloride at 68 °F is get from table 1.6 physical properties of liquid that is

σ = 2.69 × 10^{-2} N/m

so average radius = \frac{diameter}{2} =  100 µm = 100 ×10^{-6} m

now here we know relation between pressure difference and surface tension

so we can derive difference pressure as

2π×σ×r = Δp×π×r²    .....................1

here r is radius and  Δp pressure difference and σ surface tension

Δp = \frac{2 \sigma }{r}    

put here value

Δp = \frac{2*2.69*10^{-2}}{100*10^{-6}}  

Δp = 538

so the difference in pressure between the inside and outside of the droplets is 538 Pa

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