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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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For the pipe-fl ow-reducing section of Fig. P3.54, D 1 5 8 cm, D 2 5 5 cm, and p 2 5 1 atm. All fl uids are at 20 8 C. If V 1 5
bonufazy [111]

Answer:

The total force resisted by the flange bolts is  163.98 N

Explanation:

Solution

The first step is to find  the pipe cross section at the inlet section

Now,

A₁ = π /4 D₁²

D₁ =  diameter of the pipe at the inlet section

Now we insert 8 cm for D₁ which gives us A₁ = π /4 D (8)²

=50.265 cm² * ( 1 m²/100² cm²)

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Secondly, we find cross section area of  the pipe at the inlet section

A₂ = π /4 D₂²

D₂ =  diameter of the pipe at the inlet section

Now we insert 5 cm for D₁ which gives us A₁ = π /4 D (5)²

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Now,

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Where Q₁ and Q₂ are both the flow rate at the exist and inlet.

We now insert A₁V₁ for Q₁ and A₂V₂ for Q₂

So,

V₁ and V₂ are defined as the velocities at the inlet and exit

We now insert 5.0265 * 10^⁻³ m² for A₁ 5 m/s for V₁ and 1.963 * 10^⁻³ m² for A₂

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Note: Kindly find an attached copy of the part of the solution to the given question below

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Answer:

find attached

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An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500
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Answer:

Exit temperature = 32 °C

Explanation:

We are given;

Initial Pressure;P1 = 100 KPa

Cp =1000 J/kg.K = 1 KJ/kg.k

R = 500 J/kg.K = 0.5 Kj/Kg.k

Initial temperature;T1 = 27°C = 273 + 27K = 300 K

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Using ideal gas equation,

PV' = m'RT

Where m' is mass flow rate.

Thus;making m' the subject, we have;

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m' = (100 × 15)/(0.5 × 300)

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Dividing through by m', we have;

h1 + Q/m' = h2 + W/m'

h = Cp•T

Thus,

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Plugging in the relevant values, we have;

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T2 = 305 K = 305 - 273 °C = 32 °C

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