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Pachacha [2.7K]
2 years ago
11

How many moles of nitrogen, N, are in 62.0 g of nitrous oxide, N2O?​

Chemistry
1 answer:
SSSSS [86.1K]2 years ago
8 0
Answer- the estimated number is 4 moles but it actually is 3.86 moles
Explanation
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A pound of margarine costs $1.39. What is the cost of a kilogram of margarine
den301095 [7]

Answer:

              3.064 $

Solution:

First of all we will find the number of pounds contained by one Kg. Therefore, it is found that,

                                    1 Kilogram  = 2.20462 Pounds

Hence, as given,

                        1 Pound of Margarine costs  =  1.39 Dollars

So,

                  2.20462 Pounds Margarine will cost  =  X Dollars

Solving for X,

                     X  =  (2.20462 Pounds × 1.39 Dollars) ÷ 1 Pound

                     X =  3.064 Dollars

8 0
3 years ago
Which statement accurately compares the trends in atomic number and atomic mass in the periodic table?
Dmitriy789 [7]

The answer is , C. Both the atomic mass and the atomic number increase from left to right .

7 0
3 years ago
Read 2 more answers
5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buf
mr_godi [17]

Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and pK_{a} is as follows.

               pH = pK_{a} + log \frac{base}{acid}

where,     pH = 7.4 and pK_{a} = 7.21

As here, we can use the pK_{a} nearest to the desired pH.

So,      7.4 = 7.21 + log \frac{base}{acid}

             0.19 = log \frac{base}{acid}

            \frac{base}{acid} = 1.55

1 mM phosphate buffer means [HPO_{4}] + [H_{2}PO_{4}] = 1 mM

Therefore, the two equations will be as follows.

           \frac{HPO_{4}}{H_{2}PO_{4}} = 1.55 ............. (1)

  [HPO_{4}] + [H_{2}PO_{4}] = 1 mM ........... (2)        

Now, putting the value of [HPO_{4}] from equation (1) into equation (2) as follows.

             1.55[H_{2}PO_{4}] + [tex][H_{2}PO_{4}] = 1 mM

                        2.55 [H_{2}PO_{4}] = 1 mM

                             [H_{2}PO_{4}] = 0.392 mM

Putting the value of [H_{2}PO_{4}] in equation (1) we get the following.

                     0.392 mM + [HPO_{4}] = 1 mM

                          [HPO_{4}] = (1 - 0.392) mM

                              [HPO_{4}] = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.

7 0
3 years ago
How many grams of K2Cr207 are in 200 mL 0.1M
charle [14.2K]

Answer:

407grams ithink lang po

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A B C D E

Explanation:

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