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3 years ago

Fe2O3(s) + 3CO(

1 answer:

6 0

The answer is, 699g
The idea here is that you need to use the mole ratio tha exists between ferric oxide, Fe2O3, and iron metal, Fe, to determine how many moles of the latter will be produced when all the given mass of the ferric oxide reacts...

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A sample of neon occupies a volume of 752 ml at 25 0

lyudmila [28]

$V_{initial} = 752\:mL$
$T_{initial} = 25.0^0C$
converting to Kelvin
TK = TC + 273
TK = 25.0 + 273 → TK = 298.0 → $T_{initial} = 298.0\:K$
$V_{final} = ? (in\:milliliters)$
$T_{final} = 50.0^0C$
TK = TC + 273
TK = 50.0 + 273 → TK = 323.0 → $T_{final} = 323.0\:K$

By the first Law of Charles and Gay-Lussac, we have:
$\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }$

Solving:
$\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }$
$\frac{ 752 }{ 298.0 } = \frac{ V_{f} }{ 323.0 }$
Product of extremes equals product of means:
$298.0* V_{f} = 752*323.0$
$298.0 V_{f} = 242896$
$V_{f} = \frac{242896}{298.0}$
$\boxed{\boxed{V_{f} \approx 815.08\:mL}}\end{array}}\qquad\quad\checkmark$

5 0

3 years ago

Describe the structure of ammonium laurel sulfates refer to the given diagram. Your answer should include the type of bonding, t

il63 [147K]

Answer:

This ammonium laurel sulfates anion consists of a nonpolar hydrocarbon chain and a polar sulfate end group. It means it has a hydrophilic head and hydrophobic tail. There are ammonium ions, sulfate, and fatty acids present.

Lauryl sulfate has lauric acid attach to sulfate ions with carbon-sulfur bond, the fatty acid formed by the covalent bonds between C-C attached to hydrogens. Sulfur also bound to oxygen by covalent bonds. Nitrogen is surrounded by the four hydrogen atoms in the hydrophilic head.

7 0

2 years ago

Hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

MAVERICK [17]

Answer:

huh?

Explanation:

6 0

2 years ago

How to do lewis dot structure of nh3conh3

wlad13 [49]

Answer:

i'm not really sure i cant help today try a different question that i can help u with

Explanation:

3 0

2 years ago

How many milliliters of 0.128 M nitric acid are required to neutralize 11.1 mL of 0.184 M sodium hydroxide? Answer with the corr

Llana [10]

Answer:

I think 7.72 mL

11.1/184 × 128/184=7.72

4 0

2 years ago