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daser333 [38]
3 years ago
14

Even though jack works in Finance he was not always part of this career cluster. What job might jack have had before working in

finance
A assistant bank teller
B loan officer
C wedding photographer
D mortgage loan underwriter
Engineering
1 answer:
Nadya [2.5K]3 years ago
3 0
The answer is either A or B
You might be interested in
All brake lights are dimmer than normal. Technician A says that bad bulbs could be the cause. Technician B says that high resist
yarga [219]

Answer:

All Brake lights are dimmer than normal because high resistance in the brake switch could be the cause according to Technician B.

Explanation:

According to Technician A

When the bulb is faulty then no current will flow through bulb and it will be open circuit.So no light will produce in bulb .

According to Technician B

When a high resistance inserted in series  circuit the voltage across each resistance is reduced and this cause the light glow dimly.

Formula of resistance in series circuit

Rt=r1+r2+r3......

5 0
3 years ago
6.1-2. Diffusion of CO, in a Binary Gas Mixture. The gas CO2 is diffusing at stcady state through a tube 0.20 m long having a di
zzz [600]

Answer:

Heat flux of CO₂ in cgs

                 = 170.86 x 10⁻⁹ mol / cm²s

SI units

       170.86 x 10⁻⁸ kmol/m²s  

Explanation:

4 0
3 years ago
PLEASE HURRY!!!
Naily [24]

Answer:

A

Explanation:

He should get a job in engineering to see what it's like to work in the field.

3 0
2 years ago
Read 2 more answers
The concentration of carbon monoxide (co) in an exhaust gas is 1x10^4 ppmv. what is the concentration in mg/m^3 at 25 and 1 atm
Mrrafil [7]

Answer:

1.1451 x 104 (11451.13)mg/m3

Explanation:

1 ppmv is defined as one volume of a contaminant or solid(CO)(mL) in 1 x 106 volume of solvent/water.

1ppmv = 1mL/m3

Concentration in mg/m3 = volume in ppm x molecular weight x pressure(kPa)/( gas constant x temperature(K)

Molecular weight of CO = 12 + 16

= 28g/mol

Temperature = 273.15 + 25

= 298.15K

Pressure = 1 x 101.325kPa

= 101.325kPa

Ppmv = 1 x 10-4ppmv

Gas constant, R = 8.3144 L.kPa/mol.K

Concentration in mg/m3 = (1 x 104 * 28 * 101.325)/(8.3144 * 298)

= 1.1451 x 104mg/m3

= 11451.13 mg/m3

3 0
3 years ago
A liquid stream containing 52.0 mole% benzene and the balance toluene at 20.0°C is fed to a continuous single-stage evaporator a
OLga [1]

Answer:

Operating Pressure P = 793.716 mmHg

Y_Benzene y1 = 0.541

Explanation:

Given that;

liquid phase leaving the evaporator = 32.5 mole%

Equi Temp T = 99.0°C = 99 + 273.15 = 372.15 K

Now let 1 and 2 represent Benzene and Toluene respectively.

Antoine's Constant for these components are;

COMPONENETS        A                B                    C

Benzene 1             4.72583     1660.652        -1.461

Toluene  2            4.07827     1343.943         -53.773

Antoine's equation is expressed as;

Ps = 10^(A - (B/(T+C)))

Ps is in Bar and T is in Kelvin

so

P1s = 10^( 4.72583 - (1660.652/(372.15 - (-1.461)))) = 1.7617 Bar

P2s = 10^( 4.07827 - (1343.943/(372.15 - (-53.773)))) = 0.7195 Bar

now here, liquid leaving and vapor are both in equilibrium

composition of liquid leaving are;

X1 = 32.5%    = 0.325

X2 = 1 - X1 = 1 - 0.325 = 0.675

Now

Raoult's Law is expressed as;

p × y1=x1 × pis     for all components

So for Benzene ; p × y1=x1 × p1s   ------let this be equation 1

for Toluene ; p × y2=x2 × p2s   ------let this be equation 2

lets add equ 1 and 2

p × y1=x1 × p1s + p × y2=x2 × p2s

p(y1 + y2) = x1 × p1s + x2 × p2s

buy y1 + y2 = 1

therefore we substitute

p(1) = 0.325 × 1.7617 + 0.675 × 0.7195 = 1.0582 Bar

we know that 1 Bar = 750.062 mmHg

so p = 1.0582 × 750.062

p = 793.716 mmHg

Also from equation 1

p × y1=x1 × p1s

y1 = (x1 × p1s) / p

y1 = (0.325 × 1.7617) / 1.0582

y1 = 0.541

Therefore;

Operating Pressure P = 793.716 mmHg

Y_Benzene y1 = 0.541

5 0
2 years ago
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