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elixir [45]
2 years ago
15

HELLP: WHAT IS THE 10 ORGANIZATIONS OF MATTER??? WILL GIVE BRAINLEST

Chemistry
1 answer:
NARA [144]2 years ago
3 0

molecule, cell, tissue, organ, organ system, organism, population, community, ecosystem, biosphere.

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Reacting to produce hydrogen gas is a chemical property
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The action of the weather conditions in altering the color, texture, composition, or form of exposed is called ___________. (A)
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Explanation:

YES BECAUSE YE SIS YES WHEN YES=\sqrt{x} x^{2} x^{2} \neq \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right]

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2 years ago
Which of these is the second weakest of the four fundamental forces?
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It gravity

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7 0
2 years ago
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121 grams of iron (II) acetate reacts with 360 grams of lead (II) oxalate. Predict the products, write the complete reaction equ
egoroff_w [7]

Answer:

The mass of the precipitate is 100.2 grams

Explanation:

Step 1: Data given

Mass of iron (II) acetate = 121.0 grams

Mass of lead (II) oxalate = 360.0 grams

Molar mass Fe(C2H3O2)2  = 173.93 g/mol

Molar mass PbC2O4 = 295.22 g/mol

Step 2: The balanced equation

Fe(C2H3O2)2 + PbC2O4 → FeC2O4 + Pb(C2H3O2)2

Step 3: Calculate moles  iron (II) acetate

Moles = mass / molar mass

Moles = 121.0 grams /  173.93 g/mol

Moles = 0.696 moles

Step 4: Calculate moles lead (II) oxalate

Moles = 360.0 grams / 295.22 g/mol

Moles = 1.219 moles

Step 5: Calculate limiting reactant

For 1 mol iron (II) acetate we need 1 mol lead (II) oxalate to produce 1 mol Iron(II) oxalate and 1 mol Lead(II) acetate

iron (II) acetate is the limiting reactant. It will completely be consumed ( 0.696 moles). Lead (II) oxalate is in excess.  There will react 0.696 moles. There will remain 1.219 - 0.696 = 0.523 moles

Step 6: Calculate moles of iron(II) oxalate

For 1 mol iron (II) acetate we need 1 mol lead (II) oxalate to produce 1 mol Iron(II) oxalate and 1 mol Lead(II) acetate

For 0.696 moles iron(II) acetate we'll have 0.696 moles iron(II) oxalate

Step 7: Calculate mass iron(II) oxalate

Mass = moles * molar mass

Mass = 0.696 moles * 143.91 g/mol

Mass = 100.2 grams

The mass of the precipitate is 100.2 grams

7 0
3 years ago
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