X+3=2 x=?? No spamming

gulaghasi [49]

The light colored material in areas of the mood is the earliest crust on the moon.

3 0

3 years ago

The components of your sample mixture are present in trace amounts. Assuming that you will be using GC, what kind of sample intr

Oxana [17]

Answer:

Gas Chromatography (GC or GLC) is a commonly used analytic technique in many ... for quality control as well as identification and quantitation of compounds in a mixture. ... A broad variety of samples can be analyzed as long as the compounds are ... Most analytical gas chromatographs use capillary columns, where the ...

Explanation:

5 0

3 years ago

In a parallel circuit, as more resistances are added, what happens to the total circuit resistance?

grandymaker [24]

Answer:

The flow of the electrons gets slower the more resistance added

5 0

2 years ago

Consider two different types of motors. Motor A has a characteristic life of 4100 hours (based on a MTTF of 4650 hours) and a sh

Daniel [21]

Answer:B

Explanation:

Given

For motor A

Characteristic life(r)=4100 hr

MTTF=4650 hrs

shape factor(B )=0.8

For motor B

Characteristic life(r)=336 hr

MTTF=300 hr

Shape Factor (B)=3

Reliability for 100 hours

$R_a=e^{-\left ( \frac{T-r}{n}\right )B}$

$R_a=e^{-\left ( \frac{4650-4100}{100}\right )0.8}$

$R_a=e^{-4.4}=0.01227$

For B

$R_b=e^{-\left ( \frac{300-336}{100}\right )3}$

$R_b=e^{1.08}=2.944$

B is better for 100 hours

(b)For 750 hours

$R_a=e^{-0.5866}=0.55621$

$R_b=e^{0.144}=1.154$

So here B is more Reliable.

3 0

3 years ago

The elementary liquid-phase series reaction

liraira [26]

Answer:

Concentration of A: $\frac{C_{A} }{C_{Ao} } =e^{-k_{1}t }$

Concentration of B: $\frac{C_{B} }{C_{Ao} } =\frac{k_{1} }{k_{2}-k_{1} } (e^{-k_{1}t } -e^{-k_{2}t } )$

Concentration of C: $\frac{C_{C} }{C_{Ao} } =1+\frac{k_{1} }{k_{2}-k_{1} } e^{-k_{2}t } -\frac{k_{2} }{k_{2}-k_{1} } e^{-k_{1} t}$

the image shows the graphs of the three concentrations

Explanation:

We have the reaction:

A ------->k1--------->B------------->k2--------->C

Each reaction:

$r_{A} =-k_{1} C_{A} \\r_{B} =k_{1} C_{A} -k_{2} C_{B} \\r_{C} =k_{2} C_{C}$

Where Cn is the concentration of each specie (A,B,C)

The mass balance for A:

$-\frac{dC_{A} }{dt} =-r_{A} \\-\frac{dC_{A} }{dt}=k_{1} C_{A} \\-\int\limits^y_x {\frac{dC_{A} }{dt} } \,=k_{1} t\\\frac{C_{A} }{C_{Ao} } =e^{-k_{1}t }$

Where x=CAo and y=CA

The mass balance for B:

$-\frac{dC_{B} }{dt} =-r_{B} \\-\frac{dC_{B} }{dt}=k_{2} C_{B} -k_{1} C_{A} \\\frac{dC_{B} }{dt}+k_{2} C_{B}=k_{1} C_{A}\\\frac{C_{B} }{C_{Ao} } =\frac{k_{1} }{k_{2}-k_{1} } (e^{-k_{1}t }-ex^{-k_{2}t } )$

The mass balance for C:

$\frac{C_{C} }{C_{Ao} } =1-\frac{C_{A} }{C_{Ao} } -\frac{C_{B} }{C_{Ao} } \\\frac{C_{C} }{C_{Ao} }=1+\frac{k_{1} }{k_{2}-k_{1} } e^{-k_{2} t}-\frac{k_{2} }{k_{2}-k_{1} } e^{-k_{1}t }$

The maximum concentration of C is:

$C_{Cmax} =C_{Ao} (\frac{k_{2} }{k_{1} } )^{\frac{k_{2} }{k_{2}-k_{1} }} =1.6(\frac{0.01}{0.4} )^{\frac{0.01}{0.01-0.4} } =1.76mol/dm^{3}$

and the maximum time is:

$t_{max} =\frac{ln\frac{k_{2} }{k_{1} } }{k_{2}-k_{1} } =\frac{ln\frac{0.01}{0.4} }{0.01-0.4} =9.4 h$

6 0

3 years ago

X+3=2x=??No spamming​

X+3=2x=??No spamming