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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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If I want to accelerate a mass of 3 kg at 5 m/s2 then how much force should I apply?
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<u>Answer:</u> The force that must be applied is 15 N.

<u>Explanation:</u>

Force exerted on the object is defined as the product of mass of the object and the acceleration of the object.

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F=m\times a

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F = force exerted = ?

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3 years ago
2. Using the following data, calculate the average atomic mass of magnesium (give your answer to the nearest
arlik [135]

Answer:

24.32

Explanation:

From the question given above, the following data were obtained:

Isotope A:

Mass of A = 24

Abundance (A%) = 78.70%

Isotope B

Mass of B = 25

Abundance (B%) = 10.13%

Isotope C:

Mass of C = 26

Abundance (C%) = 11.17%

Average atomic mass of Mg =..?

The average atomic mass of Mg can be obtained as illustrated below:

Average atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100] + [(Mass of C × C%)/100]

Average atomic mass = [(24 × 78.70)/100] + [(25 × 10.13)/100] + [(26 × 11.17)/100]

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Therefore, the average atomic mass of magnesium (Mg) is 24.32

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