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Alex73 [517]
3 years ago
6

Determine the mass of a gold bar that has a density of 19.3 g/cm3 and is 4.72 cm high by 8.21 cm long by 3.98 cm deep.

Chemistry
1 answer:
TEA [102]3 years ago
5 0
The density of the gold is 19.3 grams/cc so each cc weighs 19.3grams. Now we can obtain the volume of gold from the given dimensions ie 4.72x8.21x3.98= 154.23 cc. So for the answer, just multiply the volume or 154.23 x 19.3= 2976.6 grams is the answer.
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5 0
1 year ago
10 points. Please help.
ratelena [41]

Answer:

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That's the Ideal Gases Law. It can be useful to solve the question.

We replace data:

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6 0
3 years ago
Read 2 more answers
Iron (III) oxide reacts with solid carbon in the followed reaction: 2Fe2O3(s) + 3C(s) → 4Fe(s) + 3CO2(g) What mass of Fe2O3 is n
Veseljchak [2.6K]

953.6 g of iron (III) oxide (Fe₂O₃)

Explanation:

We have the following chemical reaction:

2 Fe₂O₃ (s) + 3 C (s) → 4 Fe (s) + 3 CO₂ (g)

We calculate the number of moles of CO₂ by using the following formula:

pressure × volume = number of moles × gas constant × temperature

number of moles = (pressure × volume) / (gas constant × temperature)

number of moles of CO₂ = (2.1 × 100) / (0.082 × 300)

number of moles of CO₂ = 8.54 moles

Taking in account the chemical reaction we devise the following reasoning:

if         2 mole of Fe₂O₃ produces 3 mole of CO₂

then   X moles of Fe₂O₃ produces 8.54 mole of CO₂

X = (2 × 8.54) / 3 = 5.69 moles of Fe₂O₃

number of moles = mass / molar weight

mass =  number of moles × molar weight

mass of Fe₂O₃ = 5.69 × 160 = 953.6 g

Learn more about:

number of moles

brainly.com/question/14111505

#learnwithBrainly

6 0
2 years ago
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Kaylis [27]
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5. Takes place in the upper atmosphere. (A)
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3 years ago
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