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Ilia_Sergeevich [38]
3 years ago
13

When using a computer, you should not keep your:

Engineering
2 answers:
kykrilka [37]3 years ago
8 0
Your thighs should not be perpendicular and lower legs not parallel because you have either broke your knee caps backwards or your kneeling
Gre4nikov [31]3 years ago
3 0
thighs perpendicular to the floor
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The reverse water-gas shift (RWGS) reaction is an equimolar reaction between CO2 and H2 to form CO and H2O. Assume CO2 associati
klasskru [66]

Answer:

a) Check explanation for this

b)Rate law is  Rate = \frac{k_{1}k_{4}  }{k_{3}+ 2k_{4}  } [H_{2} ]

c) The rate does not depend on the concentration of CO₂

Explanation:

a) Elementary steps for the RWGS reaction:

  • Dissociative adsorption of the H₂ Molecule

                 H_{2} $\xrightarrow{\text{k1}}$H + H   (Fast process)

  • Reversible Reaction between CO₂ and H

                \[ CO_{2} + H\mathrel{\mathop{\rightleftarrows}^{\mathrm{k2}}_{\mathrm{k3}}}COOH \] (Fast Process)

  • Slow dissociation of COOH into gaseous CO and absorbed OH

                COOH $\xrightarrow{\text{k1}}$ CO + OH (Slow process)

  • Fast hydrogenation of the OH to form H₂O

                   OH + H $\xrightarrow{\text{k5}}$H_{2} O (Fast process)

b) Derivation of the rate law

We need to determine the rate law for H, OH and COOH because these are the intermediates for this reaction.

The steady state approximation is applied to a consecutive reaction with a slow first step and a fast second step (k1≪k2). If the first step is very slow in comparison to the second step, there is no accumulation of intermediate product.

Rate of consumption = Rate of production

For COOH:

Using steady state approximation

\frac{d[COOH]}{dt} = 0

k_{2} [CO_{2} ][H] = k_{3} [COOH] k_{4} [COOH]\\

[COOH] = \frac{k_{2} [CO_{2} ][H]}{k_{3}k_{4}  } \\

For H:

\frac{d[H]}{dt} = 0

k_{1}[H_{2}] = k_{2}[CO_{2} [H]+k_{5} [ OH][H]

[H]= \frac{k_{1}[H_{2}]  }{k_{5}[OH] +k_{2}[CO_{2}]}\\

For OH:

\frac{d[OH]}{dt} = 0

k_{4} [COOH] = k_{5} [OH][H]\\\k[OH] = \frac{k_{4} [COOH]}{k_{5} H}\\

The rate of the overall reaction is determined by the slowest step of the reaction. The slowest process is the dissociation of COOH

Therefore the overall rate of reaction is:

Rate = k_{4} [COOH]\\

Rate = k_{4}  \frac{k_{2} [CO_{2} ][H]}{k_{3}k_{4}  }\\Rate = k_{4}  \frac{k_{2}[CO_{2}]\frac{k_{1}[H_{2}]  }{k_{5}[OH] +k_{2}[CO_{2}]}  }{k_{3}k_{4}}\\Rate = k_{4}  \frac{k_{2}[CO_{2}]\frac{k_{1}[H_{2}]  }{k_{5}\frac{k_{4}COOH }{k_{5}H }  +k_{2}[CO_{2}]}  }{k_{3}k_{4}}

Simplifying the equation above, the rate law becomes

Rate = \frac{k_{1}k_{4}  }{k_{3}+ 2k_{4}  } [H_{2} ]

c) It is obvious from the rate law written above that the rate of the RWBG reaction does not depend on the concentration of CO₂

7 0
3 years ago
your friend's parents are worried about going over their budget for th month. Which expense would you suggest is NOT a need?
prisoha [69]

Things for entertainment aren't needs. For example, designer clothes, the new iPhone that just came out, fancy dining, that expensive shirt that has the same model somewhere cheaper, and sports cars are all wants, not needs.

Needs are like water, a shelter to live in, food, clothing, and running electricity. Water is essential for survival, since you could die of thirst in 2 days. A shelter will protect you from insects, weather, and uninvited guests. Living without food will kill you in 2 weeks, as you need food for energy and so let your body repair itself faster if you got injured. You need clothing to keep you warm and to cover your private parts. Running electricity is needed because electricity makes most of your needs easier to acquire.

I hope this helps! Good luck!

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3 years ago
What is the angle of the input
olga55 [171]
Absolute positions — latitudes and longitudes
Relative positions — azimuths, bearings, and elevation angles
Spherical distances between point locations
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For the following gear train, if the blue gear is moving at 50 rpm, what are the speeds of the other gears?
Flauer [41]

Answer:

6

Explanation:

6 teddy bears

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3 years ago
Risks in driving never begins with yourself, but with other drivers who take risks.
Ymorist [56]

False! Just saying. You could be under the influence, or just have no clue as to what you're doing.

8 0
2 years ago
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