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3 years ago

Please help! What does implementing mean?

Chemistry

1 answer:

Y_Kistochka [10]3 years ago

4 0

Implementing means to add something ie. "It's cool that you were able to implement the gummys into the root beer float"

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For the reactions below, describe the reactor system and conditions you suggest to maximize the selectivity to make the desired

alexandr1967 [171]

Answer:

hello your question lacks the required reaction pairs below are the missing pairs

Reaction system 1 :

A + B ⇒ D $-r_{1A} = 10exp[-8000K/T]C_{A}C_{B}$

A + B ⇒ U $-r_{2a} = 100exp(-1000K/T)C_{A} ^\frac{1}{2}C_{B} ^\frac{3}{2}$

Reaction system 2

A + B ⇒ D $-r_{1A} = 10exp( -1000K/T)C_{A}C_{B}$

B + D ⇒ U $-r_{2B} = 10^9exp(-10000 K/T) C_{B}C_{D}$

Answer : reaction 1 : description of the reactor system : The desired reaction which is the first reaction possess a higher activation energy and higher temperature is required to kickstart reaction 1

condition to maximize selectivity : To maximize selectivity the concentration of reaction 1 should be higher than that of reaction 2

reaction 2 :

description of reactor system : The desired reaction i.e. reaction 1 has a lower activation energy and lower temperatures is required to kickstart reaction 1

condition to maximize selectivity:

to increase selectivity the concentration of D should be minimal

Explanation:

Reaction system 1 :

A + B ⇒ D $-r_{1A} = 10exp[-8000K/T]C_{A}C_{B}$

A + B ⇒ U $-r_{2a} = 100exp(-1000K/T)C_{A} ^\frac{1}{2}C_{B} ^\frac{3}{2}$

the selectivity of D is represented using the relationship below

$S_{DU} = \frac{-r1A}{-r2A}$

hence SDu = 1/10 * $\frac{exp(-800K/T)}{exp(-1000K/T)} * C_{A} ^{0.5} C_{B} ^{-0.5}$

description of the reactor system : The desired reaction which is the first reaction possess a higher activation energy and higher temperature is required to kickstart reaction 1

condition to maximize selectivity : To maximize selectivity the concentration of reaction 1 should be higher than that of reaction 2

Reaction system 2

A + B ⇒ D $-r_{1A} = 10exp( -1000K/T)C_{A}C_{B}$

B + D ⇒ U $-r_{2B} = 10^9exp(-10000 K/T) C_{B}C_{D}$

selectivity of D

$S_{DU} = \frac{-r1A}{-r2A}$

hence Sdu = $1/10^7 * \frac{exp(-1000K/T)}{exp(-10000K/T)} *\frac{C_{A} }{C_{D} }$

description of reactor system : The desired reaction i.e. reaction 1 has a lower activation energy and lower temperatures is required to kickstart reaction 1

condition to maximize selectivity:

to increase selectivity the concentration of D should be minimal

3 0

3 years ago

2KI + Pb(NO3)2 → 2KNO3 + PbI2 Determine how many moles of KNO3 are created if 0.03 moles of KI are completely consumed.

ElenaW [278]

2:2 so the same proportion
0,03

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3 years ago

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kati45 [8]

Answer:

Explanation:

This is a correct answer i think soo

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2 years ago

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What is the result at the end of meiosis II?

VMariaS [17]

Answer:

B.) two diploid cells

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2 years ago

Calculate the ph of a 0.20 m solution of iodic acid (hio3, ka = 0.17).

saveliy_v [14]

Iodic acid partially dissociates into H+ and IO3-
Assuming that x is the concentration of H+ at equilibrium, and sine the equation says the same amount of IO3- will be released as that of H+, its concentration is also X. The formation of H+ and IO3- results from the loss of HIO3 so its concentration at equilibrium is 0.20 M - x
Ka = [H+] [IO3-] / [HIO3];
Initially, [H+] ≈ [IO3-] = 0 and [HIO3] = 0.20; 
At equilibrium [H+] ≈ [IO3-] = x and [HIO3] = 0.20 - x; 
so 0.17 = x² / (0.20 - x); 
Solving for x using the quadratic formula: 
x = [H+] = 0.063 M or pH = - log [H+] = 1.2.

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3 years ago