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jarptica [38.1K]
3 years ago
10

How many grams of AgNO3 are needed to make 250 ml of a solution that is 0.135 M

Chemistry
1 answer:
myrzilka [38]3 years ago
5 0

Answer:

( About ) 5.7 grams

Explanation:

Take a look at the attachment below for a proper explanation;

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In the industrial synthesis of hydrogen, mixtures of CO and H2 are enriched in H2 by allowing the CO to react with steam. The ch
saw5 [17]

Answer:

CO(g) + H2(g) + H2O(g) ==> CO2(g) + 2H2(g)

Explanation:

In the industry, hydrogen is prepared from water and hydrocarbons. Water gas being the major method of preparation of hydrogen industrially.

The water-gas reaction is an industrial process in which steam is passed over red-hot coke giving a gaseous mixture of carbon monoxide and hydrogen:

C + H2O(g) → CO + H2.

The mixture of CO and H2 is Futher passed through steam according to the equation:

CO(g) + H2(g) + H2O(g) ==> CO2(g) + 2H2(g) to give hydrogen and carbon dioxide.

8 0
3 years ago
The most useful ore of aluminum is bauxite, in which Al is present as hydrated oxides, Al2O3 * xH2O.
lesantik [10]

Answer:

44.7 kWh

Explanation:

Let's consider the reduction of Al₂O₃ to Al in the Bayer process.

6 e⁻ + 3 H₂O + Al₂O₃ → 2 Al + 6 OH⁻

We can establish the following relations:

  • The molar mass of Al is 26.98 g/mol.
  • 2 moles of Al are produced when 6 moles of e⁻ circulate.
  • 1 mol of e⁻ has a charge of 96468 c (Faraday's constant).
  • 1 V = 1 J/c
  • 1 kWh = 3.6 × 10⁶ J

When the applied electromotive force is 5.00 V, the energy required to produce 3.00 kg (3.00 × 10³ g) of aluminum is:

3.00 \times 10^{3} gAl.\frac{1molAl}{26.98gAl} .\frac{6mole^{-}}{2molAl}.\frac{96468c}{1mole^{-}}.\frac{5.00J}{c}.\frac{1kWh}{3.6 \times 10^{6}J} =44.7kWh

6 0
3 years ago
A 10 mm Brinell hardness indenter is used for some hardness testing measurements of a steel alloy. a) Compute the HB of this mat
Lady_Fox [76]

Explanation:

(a)  The given data is as follows.  

         Load applied (P) = 1000 kg

         Indentation produced (d) = 2.50 mm

         BHI diameter (D) = 10 mm

Expression for Brinell Hardness is as follows.

                HB = \frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}    

Now, putting the given values into the above formula as follows.

                HB = \frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}                            = \frac{2 \times 1000 kg}{3.14 \times 10 mm [D - \sqrt{((10 mm)^{2} - (2.50)^{2})}]}  

                       = \frac{2000}{9.98}                          

                       = 200

Therefore, the Brinell HArdness is 200.

(b)     The given data is as follows.

               Brinell Hardness = 300

                Load (P) = 500 kg

               BHI diameter (D) = 10 mm

             Indentation produced (d) = ?

                      d = \sqrt{(D^{2} - [D - \frac{2P}{HB} \pi D]^{2})}

                         = \sqrt{(10 mm)^{2} - [10 mm - \frac{2 \times 500 kg}{300 \times 3.14 \times 10 mm}]^{2}}

                          = 4.46 mm

Hence, the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used is 4.46 mm.

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3 years ago
11-Draw a pedigree chart for 2 parents and 4 offsprings,
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Answer:

eudora glass has been the subject for the past year and a few of its top five favorite food products are in a long term way to help the industry in a relationship and not only have the best of recipes but I am you know how I am not sure I can understand the fact ok I just don't plan to do anything for the same person as do I need a few days of good morning have to do just fine for kids to live me to be not a fan but not very much years can be so xbox or old can u do u xbox games like to watch Zoey go out

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Answer:

it increases and is perpendicular to the motion of the wave.

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3 years ago
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