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Tresset [83]
3 years ago
9

POINTS FOR HELP!!!!!

Chemistry
1 answer:
Fudgin [204]3 years ago
3 0
MgSO4, CO2, AlCl3 are the answer
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The molar heat of fusion of gold is 12.550 kJ mol–1. At its melting point, how much mass of melted gold must solidify to release
KATRIN_1 [288]

The mass of melted gold to release the energy would be  3, 688. 8 Kg

<h3>How to determine the mass</h3>

The formula for quantity of energy is given thus;

Q = n × HF

Where n represents number of moles

HF  represents  heat of fusion

To find the number of moles, we have

235.0 = n × 12.550

number of moles = \frac{235}{12. 550} = 18. 725 moles

Note that molar mass of Gold is 197g/ mol

Let's note that;

Number of moles = mass/ molar mass

Mass = number of moles × molar mass

Mass = 18. 725 × 197

Mass = 3, 688. 8 Kg

Thus, the mass of melted gold to release the energy would be  3, 688. 8 Kg

Learn more about molar heat of fusion here:

brainly.com/question/15634085

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8 0
2 years ago
Textbook mass 2000 grams volume 4000cm3 density
PIT_PIT [208]
Density = mass/volume = 2000/4000 = 0.5 grams/cm3. Hope this hopes!
8 0
3 years ago
What is the answer to 1 and 2?
ankoles [38]
You got the questions correct
4 0
3 years ago
. -9 + 6 - (-2)<br> Answers?
Kaylis [27]
The answer is -1. 2 subtraction signs next to each other for an addition sign
5 0
3 years ago
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Urea, CH4N2O (s), is manufactured from NH3 (g) and CO2 (g). H2O (l) is another product of this reaction. An experiment is starte
Katarina [22]

Answer:

a. 4.41 g of Urea

b. 1.5 g of Urea

Explanation:

To start the problem, we define the reaction:

2NH₃ (g) +  CO₂ (g) → CH₄N₂O (s)  +  H₂O(l)

We only have mass of ammonia, so we assume the carbon dioxide is in excess and ammonia is the limiting reactant:

2.6 g . 1mol / 17g = 0.153 moles of ammonia

Ratio is 2:1. 2 moles of ammonia can produce 1 mol of urea

0.153 moles ammonia may produce, the half of moles

0153 /2 = 0.076 moles of urea

To state the theoretical yield we convert moles to mass:

0.076 mol . 58 g/mol = 4.41 g

That's the 100 % yield reaction

If the percent yield, was 34%:

4.41 g . 0.34 = 1.50 g of urea were produced.

Formula is (Yield produced / Theoretical yield) . 100 → Percent yield

3 0
3 years ago
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