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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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<u>Answer:</u> The mass of cryolite produced is 51.48 kg

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

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Given mass of aluminium oxide = 12.5 kg = 12500 g    (Conversion factor:  1 kg = 1000 g)

Molar mass of aluminium oxide = 101.96 g/mol

Putting values in equation 1, we get:

\text{Moles of aluminium oxide}=\frac{12500g}{101.96g/mol}=122.6mol

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Putting values in equation 1, we get:

\text{Moles of NaOH}=\frac{55400g}{40g/mol}=1389mol

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Al_2O_3(s)+6NaOH(l)+12HF(g)\rightarrow 2Na_3AlF_6+9H_2O(g)

By Stoichiometry of the reaction:

1 mole of aluminium oxide reacts with 6 moles of sodium hydroxide and 12 moles of HF.

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As, given amount of NaOH and HF is more than the required amount. So, they are considered as an excess reagent.

Thus, aluminium oxide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of aluminium oxide produces 2 moles of cryolite

So, 122.6 moles of aluminium oxide will produce = \frac{2}{1}\times 122.6=245.2mol of cryolite

Now, calculating the mass of cryolite by using equation 1:

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Moles of cryolite = 245.2 mol

Putting values in equation 1, we get:

245.2mol=\frac{\text{Mass of cryolite}}{209.94g/mol}\\\\\text{Mass of cryolite}=(245.2mol\times 209.94g/mol)=51477.3g

Converting this into kilograms, we use the conversion factor:

1 kg = 1000 g

So, 51477.3 g\times (\frac{1kg}{1000g})=51.48kg

Hence, the mass of cryolite produced is 51.48 kg

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