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VMariaS [17]
3 years ago
11

The rate of disappearance of HBr in the gas phase reaction 2HBr(

Chemistry
1 answer:
Iteru [2.4K]3 years ago
3 0
The correct answer would be 0.1505 M/s. Given the rate of disapperance of HBr, we can easily calculate the rate of apearance of the products by looking at the coefficients of the substances in the reaction. In this case, for every two moles of HBr, 1 mole of Br2 is being produced so the reaction rate would be:

- 1/2 (r(HBr)) = rBr2
- 1/2 (-0.301) = rBr2
Rate of appearance of Br2 = 0.1505 M/s
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If 294 grams of FeS2 is allowed ti react with 176 grams of O2 according to the following equation how many grams of Fe2O3 are pr
podryga [215]
 <span>You must balance your equation correctly.
Here is your answer: 

294gFeS2 x 1molFeS2/119.99 x 11mols O2/4mols FeS2--> 6.738mol O2 

176gO2 x 1mol O2/32gO2 x 4mols FeS2/11mol FeS2--> 2mols FeS2 
Now choose the molecule with the lowest amount (Limiting Reagent) 

2molsFeS2 x 2molsFe2O3/4molsFeS2 x 159.7g
159.7g Fe2O3 grams produced.</span>
6 0
3 years ago
Read 2 more answers
2. What mass of water absorbs 6700 J of heat to raise the temperature from 283K to 318K?​
Alexxandr [17]

Answer:

Q = mcT ...you can either substitute the molar heat capacity of water in the place of c or the specific heat capacity of water.

Explanation:

7 0
2 years ago
Water (with density of 1000 kg/m3) with the mass flowrate of 10 kg/sec is flowing into an empty tank. The outlet volumetric flow
Montano1993 [528]

Explanation:

Apply the mass of balance as follows.

   Rate of accumulation of water within the tank = rate of mass of water entering the tank - rate of mass of water releasing from the tank

         \frac{d}{dt}(\rho V) = 10 - \rho \times (0.01 h)

      \rho A_{c} \frac{dh}{dt} = 10 - (0.01) \rho h

   \frac{dh}{dt} + \frac{0.01 \rho h}{\rho A_{c}} = \frac{10}{\rho A_{c}}

          [/tex]\frac{dh}{dt} + \frac{0.01}{0.01}h[/tex] = \frac{10}{\rho A_{c}}

                       A_{c} = 0.01 m^{2}

              \frac{dh}{dt} + h = 1

                  \frac{dh}{dt} = 1 - h

               \frac{dh}{1 - h} = dt  

                \frac{ln(1 - h)}{-1} = t + C      

Given at t = 0 and V = 0  

                         A \times h = 0  

 or,                     h = 0

                 -ln(1 - h) = t + C

Initial condition is -ln(1) = 0 + C

                                C = 0  

                So,   -ln(1 - h) = t

or,                      t = ln (\frac{1}{1 - h})  ........... (1)

(a)    Using equation (1) calculate time to fill the tank up to 0.6 meter from the bottom as follows.

                    t = ln (\frac{1}{1 - h})  

                     t = ln (\frac{1}{1 - 0.6})  

                        = ln (\frac{1}{0.4})

                        = 0.916 seconds

(b)   As maximum height of water level in the tank is achieved at steady state that is, t = \infty.  

                    1 - h = exp (-t)

                    1 - h = 0  

                         h = 1

Hence, we can conclude that the tank cannot be filled up to 2 meters as maximum height achieved is 1 meter.

                 

8 0
3 years ago
What is the mass in grams of 0.7350 moles of sodium?
Alik [6]

Answer:

Explanation:

1 mol of sodium = 23 grams (use the number on your periodic table).

0.7350 mol sodium = x

Cross multiply

1*x = 0.7350 * 23

x = 16.905

You will get slightly less than this, depending on your periodic table. But the method will be the same.

4 0
3 years ago
HELP<br>12 (a crossword that had 7 spaces)
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JJ Thompson discovered electrons
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