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Anastaziya [24]
4 years ago
6

Two iron oxide samples are given to you where one is red and the other is black. You perform a chemical analysis and you find th

at the red sample has a Fe/O mass ratio of 2.327 and the black has a Fe/O mass ratio of 3.491. You suspect the red sample is simple rust or Fe2O3. What is the chemical formula for the black sample?
Chemistry
1 answer:
kupik [55]4 years ago
7 0

Answer:

Chemical formula for the black iron oxide sample is FeO

Explanation:

<u>Given:</u>

Mass ratio of black iron oxide sample, Fe/O = 3.491

<u>To determine:</u>

The chemical formula

<u>Calculation:</u>

The mass ratio is Fe:O = 3.491 : 1

Mass of Fe = 3.491 g

Mass of O = 1.000 g

Atomic mass of Fe =55.85 g/mol

Atomic mass of O = 16.00 g/mol

Moles\ Fe = \frac{3.491g}{55.85g/mol} =0.625\\\\Moles\ O = \frac{1.000g}{16.00g/mol} =0.0625\\

Therefore, the molar ratio of Fe:O = 1:1

Hence, chemical formula is FeO

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How many grams of liquid water are produced when 60 grams of ice melt?
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Answer: 60 Grams of water are produced, because you can't get more than the amount that was in the original ice cube

Explanation:

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3 years ago
if 0.582 moles of zinc reacts with excess lead(IV) sulfate how many grams are zinc sulfate would be produced? can someone help m
love history [14]
Balanced equation:
Pb(SO₄)₂ + 2 Zn → 2 ZnSO₄ + Pb
From the equation we found that 2 moles of Zn form 2 moles of ZnSO₄ 
so 0.582 mole of Zn will produce 0.582 ZnSO₄
Molar mass of ZnSO₄ = 161.47 g/mole
Mass of ZnSO₄ formed = 161.47 x 0.582 = 94.0 grams 
7 0
3 years ago
Calculate the molar mass of carbon tetrafluoride (CF4) in grams per mole, rounding to proper significant figures, if mc= 12.01 u
Nata [24]

Answer:

molar mass of carbon tetrafluoride (CF4) is

(12.01 × 1 ) + ( 4 × 19.00)

= 12.01 + 76

= 88.01u

= 88u

Hope this helps

3 0
3 years ago
Read 2 more answers
Please someone help I’m really confused
NARA [144]

Answer:10-3

Explanation:

7 0
3 years ago
consider the reaction between calcium oxide and carbon dioxide: cao ( s ) + co 2 ( g ) → caco 3 ( s ) a chemist allows 14.4 g of
lana [24]

Answer:

Explanation:

Given data:

Mass of calcium oxide = 14.4 g

Mass of carbon dioxide = 13.8 g

Actual yield of calcium carbonate = 19.4 g

Mass of calcium carbonate produced = ?

Limiting reactant = ?

Percent yield = ?

Chemical equation:

CaO + CO₂  → CaCO₃

Number of moles of CaO:

Number of moles of CaO = Mass /molar mass

Number of moles of CaO = 14.4 g / 56.1g/mol

Number of moles of CaO = 0.26 mol

Number of moles of CO₂:

Number of moles of CO₂= Mass /molar mass

Number of moles of CO₂ = 13.8 g / 44 g/mol

Number of moles of CO₂ = 0.31 mol

Now we will compare the moles of CaCO₃ with CO₂ and CaO.

                  CaO           :              CaCO₃

                    1               :                 1

                 0.26           :            0.26

                  CO₂           :                CaCO₃

                  1                 :                 1

                 0.31            :               0.31

The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.

Limiting reactant:

CaO

Theoretical yield:

Mass of CaCO₃ = moles × molar mass

Mass of  CaCO₃ = 0.26 mol × 100 g/mol

Mass of  CaCO₃ =  26 g

Percent yield:

Percent yield = Actual yield / theoretical yield × 100

Percent yield = 19.4 g/ 26 g× 100

Percent yield = 74.6 %

3 0
3 years ago
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