A chemical engineer can clearly see from this kind of test if a substance stays in a system and builds up or if it just passes through.
<h3>What is a chemical engineer?</h3>
- Processes for manufacturing chemicals are created and designed by chemical engineers.
- To solve issues involving the manufacture or usage of chemicals, fuel, medications, food, and many other goods, chemical engineers use the concepts of chemistry, biology, physics, and math.
- A wide range of sectors, including petrochemicals and energy in general, polymers, sophisticated materials, microelectronics, pharmaceuticals, biotechnology, foods, paper, dyes, and fertilizers, have a significant demand for chemical engineers.
- Chemical engineering is undoubtedly difficult because it requires a lot of physics and math, as well as a significant number of exams at the degree level.
To learn more about chemical engineer, refer to:
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Answer:
Explanation: Here it is: 67 Hope that helps! :)
Answer:
the rate of heat loss by convection across the air space = 82.53 W
Explanation:
The film temperature

to kelvin = (5 + 273)K = 278 K
From the " thermophysical properties of gases at atmospheric pressure" table; At
= 278 K ; by interpolation; we have the following
→ v 13.93 (10⁻⁶) m²/s
→ k = 0.0245 W/m.K
→ ∝ = 19.6(10⁻⁶)m²/s
→ Pr = 0.713

The Rayleigh number for vertical cavity

= 
= 

For the rectangular cavity enclosure , the Nusselt number empirical correlation:





h = 1.99 W/m².K
Finally; the rate of heat loss by convection across the air space;
q = hA(T₁ - T₂)
q = 1.99(1.4*0.96)(20-(-10))
q = 82.53 W
<h2>Answer:</h2>
<h3>Required Answer is as follows :-</h3>
- 3/4 = (⅓ × m₁/Volume × V²)/(⅓ × m₂/Volume) × (V/2)²
- ¾ = m₁/Volume × (V)²/m₂/Volume × (V/2)²
- ¾ = m₁ × (V)²/m₂ × (V²/4)
<h3>Now,</h3>
- ∆m = m₂ - m₁
- ∆m = 16m₂/3 - m₁ = 13m₁/3
- Ratio = (13m₁/3)/ m₁
- Ratio = 13/3
Ratio = 13:3
<h3>Know More :-</h3>
Mass => It is used to measure it's resistance to its acceleration. SI unit of mass is Kg.
Answer:
Equation:

Supose we have 1 kmol of ethane, according to the equation we should have 7/2 kmol of oxygen, witch corresponds to a number of mols of air of:

and air is 0.233*32 + 0.767*28 = 28.9kg/kmol, the mass ratio is: