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svp [43]
3 years ago
14

Make the following conversion. 60 dkm = _____ km 6,000 0.060 60,000 0.60

Chemistry
1 answer:
natulia [17]3 years ago
8 0

Answer:

0.6

Explanation:

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In a hydrogen fuel cell, hydrogen gas and oxygen gas are combined to form water. Write the balanced chemical equation describing
xxTIMURxx [149]

<u>Answer:</u> The chemical reaction is given below.

<u>Explanation:</u>

A fuel cell is defined as the electrochemical cell which converts the chemical energy of a fuel (often used hydrogen) and an oxidizing agent (often used oxygen) into electrical energy via a pair of redox reactions.

The reactions which occur in hydrogen-oxygen fuel cell are:

At cathode:  H_2+2OH^-\rightarrow 2H_2O+2e^-

At anode:  \frac{1}{2}O_2+H_2O+2e^-\rightarrow 2OH^-

Net reaction:  H_2+\frac{1}{2}O_2\rightarrow H_2O

Thus, the chemical reaction is given above.

5 0
4 years ago
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How many moles are present in 57.4 grams of sulfur dioxide (SO2)?
vlabodo [156]

Answer:

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Densidad 26 288 kg/m³; 0,0026 g/cm³

Masa molar 64,0638 g/mol

Punto de fusión 198 K (−75 °C)

Punto de ebullición 263 K (−10 °C)

6 0
3 years ago
Please help :( due tomorrow
olganol [36]
1. 2650000
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4 0
2 years ago
A mixture of A and B is capable of being ignited only if the mole percent of A is 6 %. A mixture containing 9.0 mole% A in B flo
atroni [7]

Explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

             Weight = 0.09 \times 16.04 + 0.91 \times 29

                          = 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

                    \frac{800}{27.8336}

                     = 28.74 kmol/hr

Now, applying component balance as follows.

                0.09 \times 28.74 + 0 \times F_{B} = 0.06F_{p}

                   F_{p} = 43.11 kmol/hr

                   F_{A} + F_{B} = F_{p}

                          F_{B} = 43.11 - 28.74

                                      = 14.37 kmol/hr

So, mass flow rate of pure (B), is F_{B} = 14.37 \times 29

                                                    = 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

             Molecular weight of product stream = Mol. weight \times 43.11 kmol/hr

                                  = 0.06 \times 16.04 + 0.94 \times 29

                                  = 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of O_{2} into the product stream is as follows.

                    0.21 \times 0.94 \times 43.11

                      = 8.5099 kmol/hr \times 329 g/mol

                      = 272.31 kg/hr

Therefore, calculate the mass % of O_{2} into the stream as follows.

                 \frac{272.31}{1216.67} \times 100

                     = 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of O_{2} in the product gas is 22.38%.

4 0
3 years ago
Calculate the change in pH for each buffer upon the addition of 1.0 mL of 1.00MHCl. Express your answers using two decimal place
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1.9 mL. en r t tu y y y
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