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3 years ago

15

Acredit report summarizes a person's Acredit score is measure of a person's as a borrower a factor that contributes to a person'

s credit score

1 answer:

krok68 [10]3 years ago

4 0

Answer:

1) credit activity

2) trustworthiness

3) payment history

Explanation:

just checked on edg

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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

How do i play Fortnite on controller?

neonofarm [45]

Answer:

use Bluetooth if you have an unwired controller, but If its wired just hook it up.

4 0

2 years ago

Read 2 more answers

Whats the purpose of the keyway

Nata [24]

Answer:

abrir candados y abrir puertas

Explanation:

4 0

2 years ago

Flow and Pressure Drop of Gases in Packed Bed. Air at 394.3 K flows through a packed bed of cylinders having a diameter of 0.012

devlian [24]

The pressure drop of air in the bed is 14.5 kPa.

<u>Explanation:</u>

To calculate Re:

$R e=\frac{1}{1-\varepsilon} \frac{\rho q d_{p}}{\mu}$

From the tables air property

$\mu_{394 k}=2.27 \times 10^{-5}$

Ideal gas law is used to calculate the density:

ρ = $\frac{2.2}{2.83 \times 10^{-3} \times 394.3}$

ρ = 1.97 Kg / $m^{3}$

ρ = $\frac{P}{RT}$

R = $\frac{R_{c} }{M}$ = 8.2 × $10^{-5}$ / 28.97× $10^{-3}$

R = 2.83 × $10^{-3}$ $m^{3}$ atm / K Kg

q is expressed in the unit m/s

$q=\frac{2.45}{1.97}$

q = 1.24 m/s

Re = $\frac{1}{1-0.4} \frac{1.97 \times 1.24 \times 0.0127}{2.27 \times 10^{-5}}$

Re = 2278

The Ergun equation is used when Re > 10,

$\frac{\Delta P}{L}=\frac{180 \mu}{d_{p}^{2}} \frac{(1-\varepsilon)^{2}}{\varepsilon^{3}} q+\frac{7}{4} \frac{\rho}{d_{p}} \frac{(1-\varepsilon)}{\varepsilon^{3}} q^{2}$

$\frac{\Delta P}{L}=\frac{180 \times 2.27 \times 10^{-5}}{0.0127^{2}} \frac{(1-0.4)^{2}}{0.4^{3}} 1.24$ $+\frac{7}{4} \frac{1.97}{0.0127} \frac{(1-0.4)}{0.4^{3}} 1.24^{2}$

= 4089.748 Pa/m

ΔP = 4089.748 × 3.66

ΔP = 14.5 kPa

4 0

3 years ago

A bar having a length of 5 in. and cross-sectional area of 0. 7 in.2 is subjected to an axial force of 8000 lb. If the bar stret

andrew11 [14]

The modulus of elasticity is 28.6 X 10³ ksi

<u>Explanation:</u>

Given -

Length, l = 5in

Force, P = 8000lb

Area, A = 0.7in²

δ = 0.002in

Modulus of elasticity, E = ?

We know,

Modulus of elasticity, E = σ / ε

Where,

σ is normal stress

ε is normal strain

Normal stress can be calculated as:

σ = P/A

Where,

P is the force applied

A is the area of cross-section

By plugging in the values, we get

σ = $\frac{8000 X 10^-^3}{0.7}$

σ = 11.43ksi

To calculate the normal strain we use the formula,

ε = δ / L

By plugging in the values we get,

ε = $\frac{0.002}{5}$

ε = 0.0004 in/in

Therefore, modulus of elasticity would be:

$E = \frac{11.43}{0.004} \\\\E = 28.6 X 10^3 ksi$

Thus, modulus of elasticity is 28.6 X 10³ ksi

6 0

2 years ago