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3 years ago

A(n) _________ light bulb uses the most energy and is not good to use for conserving energy.

1 answer:

3 0

LED light bulb is more energy efficient than CFL bulb simply because they are based on better technology. LED light bulbs use much lesser energy in comparison to the traditional technology.

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Helium is listed as a noble gas true or false

Nonamiya [84]

True; Heilum is in Group 18 which are called “Noble Gases”

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4 years ago

Fluorine gas is placed in contact with calcium metal at high temperatures to produce calcium fluoride powder. What is the formul

Rudik [331]

F₂(g) + Ca(s) → CaF₂(s)

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4 years ago

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Diego runs away from zombies at 11 km/h for 3 h.<br> How far does he go?

aleksley [76]

Answer:

33 km in 3 hours

11/1= 11

11*3=33

Explanation:

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3 years ago

Caleb wants to find out which type of floor material is best for bouncing tennis balls the highest. Which statement best describ

goldenfox [79]

Option 2 seems to be the best answer

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3 years ago

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A mixture of A and B is capable of being ignited only if the mole percent of A is 6 %. A mixture containing 9.0 mole% A in B flo

atroni [7]

Explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

Weight = $0.09 \times 16.04 + 0.91 \times 29$

= 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

$\frac{800}{27.8336}$

= 28.74 kmol/hr

Now, applying component balance as follows.

$0.09 \times 28.74 + 0 \times F_{B} = 0.06F_{p}$

$F_{p}$ = 43.11 kmol/hr

$F_{A} + F_{B} = F_{p}$

$F_{B}$ = 43.11 - 28.74

= 14.37 kmol/hr

So, mass flow rate of pure (B), is $F_{B}$ = $14.37 \times 29$

= 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

Molecular weight of product stream = $Mol. weight \times 43.11 kmol/hr$

= $0.06 \times 16.04 + 0.94 \times 29$

= 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of $O_{2}$ into the product stream is as follows.

$0.21 \times 0.94 \times 43.11$

= $8.5099 kmol/hr \times 329 g/mol$

= 272.31 kg/hr

Therefore, calculate the mass % of $O_{2}$ into the stream as follows.

$\frac{272.31}{1216.67} \times 100$

= 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of $O_{2}$ in the product gas is 22.38%.

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3 years ago