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kipiarov [429]
1 year ago
11

a sample of gold required 2.1200j of heat to melt it from room temperature, 22.0 degrees celsius to its melting point, 1064.4 de

grees celsius. How many pounds of gold were in the sample?
Chemistry
1 answer:
ElenaW [278]1 year ago
7 0

Mass of Gold = 267.165 ×  0.01552494829

⇒ 4.1477228099

The amount of heat(q) required to raise m grams of a substance-specific C from T1 to T2 is given by

q=m C (T2-T1) ........1

Given : q= 2.1200 J

the initial temperature of gold, T1 = 22.0Celcius

the final temperature of gold, T2 = 1064.4Celcius

specific heat of gold = 0.131

putting values in eq 1:

⇒ 2.1200 = m × 0.131 × (1064.4-22)

⇒ 2.1200 = m × 0.131 × 1042.4

⇒ 2.1200 / 136.5544

⇒ 0.01552494829

Since 1g= 0.01552494829 Pounds

Mass of Gold = 267.165 ×  0.01552494829

⇒ 4.1477228099

Learn more about temperature here: brainly.com/question/11464844

#SPJ9

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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.

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