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mars1129 [50]
3 years ago
10

0.1005 liters is the same as: A. 0.0001005 cm3 B.0.1005 cm3 C.100.5 cm3 D.0.01005 cm3 and A. 0.01005 mL B. 0.1005 mL C. 0.000100

5 mL D. 100.5 mL
Chemistry
2 answers:
Andreyy893 years ago
4 0

I think it would be C.100.5cm or D.100.5ml hope that helps


Alecsey [184]3 years ago
4 0

The correct answers are: C. 100.5cm³ and D. 100.5 mL


FIRST.

Cubic centimeter (cm³) is the volume made by a cube that is 1 centimeter on each side.

<em>C. 100.5cm³</em>

1L=1000cm^3 \\ \\ So: \\ \\ 0.1005L=1000\times 0.1005cm^3=\boxed{100.5cm^3}


SECOND.

<em>D. 100.5 mL</em>

1 cubic centimeter is equal to 1mL, which is one-thousandth of a liter. So:


1L=1000mL \\ \\ So: \\ \\ 0.1005L=1000\times 0.1005mL=\boxed{100.5mL}

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2Al+ Fe203 Al203 +2Fe
leonid [27]

Answer:

229 g Al₂O₃; 243 g Fe₂O₃

Explanation:

We have the masses of two reactants, so this is a <em>limiting reactant problem</em>.  

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.  

Step 1. <em>Gather all the information</em> in one place with molar masses above the formulas and masses below them.  

M_r:      26.98    159.69    101.96

              2Al   +   Fe₂O₃ ⟶ Al₂O₃ + 2Fe

Mass/g:  121          601

===============

Step 2. Calculate the <em>moles of each reactant </em>

Moles of Al         = 121 × 1/26.98

Moles of Al         = 4.485 mol Al

Moles of Fe₂O₃  = 601× 1/159.69

Moles of Fe₂O₃  = 3.764 mol Fe₂O₃

===============

Step 3. Identify the <em>limiting reactant</em>  

Calculate the moles of Al₂O₃ we can obtain from each reactant.  

<em>From Al </em>

The molar ratio is 1 mol Al₂O₃:2 mol Al

Moles of Al₂O₃ = 4.485 × 1/2

Moles of Al₂O₃ = 2.242 mol Al₂O₃

<em>From Fe₂O₃</em>:

The molar ratio is 1 mol Al₂O₃:1 mol Fe₂O₃

Moles of Al₂O₃ = 3.764 × 1/1

Moles of Al₂O₃ = 3.764 mol Al₂O₃

The <em>limiting reactant</em> is Al because it gives the smaller amount of Al₂O₃.

The <em>excess reactant</em> is Fe₂O₃.

===============

Step 4. Calculate the <em>mass of Al₂O₃ formed </em>

Mass of Al₂O₃ = 2.242 × 101.96

Mass of Al₂O₃ = 229 g Al₂O₃

===============

Step 5. Calculate the <em>moles of Fe₂O₃ reacted </em>

The molar ratio is 1 mol Fe₂O₃:2 mol Al:

Moles of Fe₂O₃ = 4.485 × ½

Moles of Fe₂O₃ = 2.242 mol Fe₂O₃

===============

Step 6. Calculate the <em>moles of Fe₂O₃ remaining </em>

Moles remaining = original moles – moles used

Moles remaining = 3.764 – 2.242

Moles remaining = 1.521 mol Fe₂O₃

==============

Step 7. Calculate the <em>mass of Fe₂O₃ remaining </em>

Mass of Fe₂O₃ = 1.521 × 159.69/1

Mass of Fe₂O₃ = 243 g Fe₂O₃

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What tests would you use to identify a solution of fecl3
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A flame test of a colorless solution gives a bright yellow color. When reacted with AgNO, a white precipitate forms that dissolves when HNO, is added.

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