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SIZIF [17.4K]
3 years ago
15

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

235.0 kJ of energy?
Chemistry
2 answers:
KIM [24]3 years ago
7 0

Answer:

the answer is d.

Explanation:

Excuse me while I go kill Thanos.

GuDViN [60]3 years ago
5 0
<span>The total heat released, Q, equals number of moles, n, times molar heat of fussion, Hf. this is: Q = n * Hf. You kno Q and Hf, so you can find n as n = Q/Hf = 235.0 kJ / 12.55 kJ/mol = 18.72 mol. Now you can pass that to mass using the atomic mass of Au, which is 197 g/mol => mass = 18.73 mol * 197 g/mol = 3688.9 grams.</span>
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A science student wants to use distillation to separate a mixture of two
statuscvo [17]

Answer:

Boiling point

Explanation:

Distillation is one of the most widely used separation technique in chemistry. It is used to separate a mixture of liquid substances with different boiling point. Hence, the basis of the separation is BOILING POINT DIFFERENCE.

In the procedure, the liquid substances are heated until they turn gaseous, which they do at different times considering their different boiling points. The separated components are then converted back to liquid states in a process called CONDENSATION.

6 0
3 years ago
Copper is an example of a(n)
expeople1 [14]

copper is not a stone, copper is not a consumption resource, and copper is not renewable, that means that :

d. nonrenewable resource

Explanation:

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7 0
3 years ago
Read 2 more answers
You have a solution of 600 mg of caffeine dissolved in 100 mL of water. The partition coefficient for aqueous caffeine extracted
klio [65]

Answer:

159 mg caffeine is being extracted in 60 mL dichloromethane

Explanation:

Given that:

mass of caffeine in 100 mL of water =  600 mg

Volume of the water = 100 mL

Partition co-efficient (K) = 4.6

mass of caffeine extracted = ??? (unknown)

The portion of the DCM = 60 mL

Partial co-efficient (K) = \frac{C_1}{C_2}

where; C_1= solubility of compound in the organic solvent and C_2 = solubility in aqueous water.

So; we can represent our data as:

K=(\frac{A_{(g)}}{60mL} ) ÷ (\frac{B_{(mg)}}{100mL} )

Since one part of the portion is A and the other part is B

A+B = 60 mL

A+B = 0.60

A= 0.60 - B

4.6=(\frac{0.6-B(mg)}{60mL} ) ÷ (\frac{B_{(mg)}}{100mL})

4.6 = \frac{(\frac{0.6-B(mg)}{60mL} )}{(\frac{B_{(mg)}}{100mL})}

4.6 × (\frac{B_{(mg)}}{100mL}) = (\frac{0.6-B(mg)}{60mL} )

4.6 B *\frac{60}{100} = 0.6 - B

2.76 B = 0.6 - B

2.76 + B = 0.6

3.76 B = 0.6

B = \frac{0.6}{3.76}

B = 0.159 g

B = 159 mg

∴ 159 mg caffeine is being extracted from the 100 mL of water containing 600 mg of caffeine with one portion of in 60 mL dichloromethane.

4 0
3 years ago
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irina1246 [14]

A. Breaking a Cl-Br bond

6 0
2 years ago
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nexus9112 [7]
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