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Delvig [45]
2 years ago
11

A rotameter calibration curve (flow rate versus float position) obtained using a liquid is mistakenly used to measure a gas flow

rate. Would you expect the gas rate determined in this manner to be too high or too low?
Chemistry
1 answer:
Veronika [31]2 years ago
5 0

Answer:

I would expect the gas rate determined in this manner to be too low

Explanation:

A Rotameter can be designed to respond to the sensitivity of density, velocity, to measure the flow rate of liquid or gas enclosed in a tube. Liquids are denser than gas, and since the gas rate to be determined needed to respond to the velocity head alone of the rotameter so as to bring the forces in the tube equilibrium. Knowing if there is no flow, then the float would remain at the bottom, so gas has to flow at a higher rate compared to the liquid so the float would be in a similar position making it easier to measure the flowrate. This leaves the gas rate to be determined too low.

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In the reaction N2 + 3H2 → 2NH3, what is the mole ratio of hydrogen to nitrogen?
uranmaximum [27]

Answer: The mole ratio of hydrogen to nitrogen is 3 mole: 1 mole, 3:1

Explanation:

•Mole ratios are determined using the coefficients of the substances in the balanced chemical equation. •Each coefficient represents the number of mole of each substance in the chemical reaction.

•The mole ratio can be determined by first writing out a balanced chemical equation for the reaction.

For this reaction the balanced chemical equation is

N2(g) + 3H2(g) ----> 2NH3(g)

1mol:3mol : 2mol

From the equation we can see that 1 mole of N2(g) reacts with 3 moles of H2(g) or 3 moles of H2(g) react with 1 mole of N2(g) to produce 2 moles of NH3(g).

Therefore, the mole ratio of hydrogen to nitrogen is 3 mole: 1 mole, 3:1

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3 years ago
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AURORKA [14]

Answer: yes

Explanation: no

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2 years ago
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"A sphere of radius 0.50 m, temperature 27oC, and emissivity 0.85 is located in an environment of temperature 77oC. What is the
Aleksandr-060686 [28]

Explanation:

It is known that formula for area of a sphere is as follows.

                     A = 4 \pi r^{2}

                        = 4 \times 3.14 \times (0.50 m)^{2}

                        = 3.14 m^{2}

    T_{a} = (27 + 273.15) K = 300.15 K

          T = (77 + 273.15) K = 350.15 K

Formula to calculate the net charge is as follows.

             Q = esA(T^{4} - T^{4}_{a})

where,    e = emissivity = 0.85

               s = stefan-boltzmann constant = 5.6703 \times 10^{-8} Wm^{-2} K^{-4}

                A = surface area

Hence, putting the given values into the above formula as follows.

                 Q = esA(T^{4} - T^{4}_{a})

                     = 0.85 \times 5.6703 \times 10^{-8} Wm^{-2} K^{-4} \times 3.14 \times ((350.15)^{4} - (300.15)^{4})

                     = 1046.63 W

Therefore, we can conclude that the net flow of energy transferred to the environment in 1 second is 1046.63 W.

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