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3 years ago

How many moles in 843.211 grams of c6h12o6?

1 answer:

3 0

Molar mass C6H12O6 = 12 x 6 + 1 x 12 + 16 x 6 = 180 g/mol

1 mole ------------ 180 g
( moles ) --------- 843.211 g

moles = 843.211 x 1 / 180

moles = 843.211 / 180

= 4,684 moles of C6H12O6

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750.0g of water that was just boiled (heated to 100.0/c loses 78.45KJ of heat as it cools.What is the final temperature of water

Iteru [2.4K]

Mass=m=750g=0.75kg
Q=78.45\times 10^3J=78450J
T_i=100°C
T_f=?
Specific heat capacity=c=4200J/kg°C

According to thermodynamics

$\\ \tt\bull\rightarrowtail Q=mc\Delta T$

$\\ \tt\bull\rightarrowtail Q/mc=\Delta T$

$\\ \tt\bull\rightarrowtail \Delta T=\dfrac{78450}{0.75(4200)}$

$\\ \tt\bull\rightarrowtail \Delta T=24.9°C$

$\\ \tt\bull\rightarrowtail T_i-T_f=24.9$

$\\ \tt\bull\rightarrowtail T_f=100-24.9=76.1°C$

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2 years ago

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

How many cans of dog food must Sam buy for his 3 dogs if each dog eats 2 cans per day? Sam only shops once a week.

sergey [27]

To find the answer you can set up an equation like this: 3(x)(d)
x would stand for number of cans each dog eats, while d stands for the number of days in a week. 3(2)(7)=42 Sam would have to purchase 42 cans of dog food each week.

7 0

3 years ago

Question 2 (2 points) Question 2 Unsaved Why does a blue shirt look blue? Choose the BEST explanation. Question 2 options: All w

Mazyrski [523]

1 . Each color has a different wavelength allowing the eye to see it.2 . The shirt reflects the blue wavelengths.
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3 years ago

A mixture of 454 kg of applesauce at 10 degrees Celsius is heated in a heat exchanger by adding 121300 kJ. Calculate the outlet

stira [4]

Explanation:

The given data is as follows.

Mass of apple sauce mixture = 454 kg

Heat added (Q) = 121300 kJ

Heat capacity ( $C_{p}$ ) of apple sauce at $32.8^{o}C$ = 4.0177 $kJ/kg^{o}C$