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vredina [299]
3 years ago
15

How many formula units are in 13.2 grams of Fe2O3?

Chemistry
1 answer:
PSYCHO15rus [73]3 years ago
4 0

Answer:

2.651g

Explanation:

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When did oxygen first appear in earth's atmosphere??
ad-work [718]
<span> the atmosphere holds about 21 per cent oxygen. Over the Earth’s 4.6 billion year history, oxygen did not appear in the atmosphere until perhaps about 2.5 billion years ago. Since then, oxygen levels have fluctuated in tandem with global geological and biological events, such as mass extinctions.</span>
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A 250ml glass of orange juice contains 22 grams of sugar. how much sugar is in a 2 liter (2.500ml) bottle of orange juice?
Maslowich

220 grams of sugar would be in 2 liters of orange juice

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Select the answers below that represent the ionic compounds in this list: na2so4 h2so4 ch3-ch3 n2o4 cacl2 ca3(po4)2
yarga [219]

Ionic compounds are those compounds which form an ion or dissociates into ions when placed in a solution. These compounds are made up of a metal and non metal. From the choices, here are the ionic compounds:

na2so4

h2so4

cacl2

<span>ca3(po4)2</span>

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4 years ago
What explains the relationship among elements, compounds &amp; mixtures
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3 0
3 years ago
Suppose 6.63g of zinc bromide is dissolved in 100.mL of a 0.60 M aqueous solution of potassium carbonate. Calculate the final mo
Alenkasestr [34]

Answer:

[Zn²⁺] = 4.78x10⁻¹⁰M

Explanation:

Based on the reaction:

ZnBr₂(aq) + K₂CO₃(aq) → ZnCO₃(s) + 2KBr(aq)

The zinc added produce the insoluble ZnCO₃ with Ksp = 1.46x10⁻¹⁰:

1.46x10⁻¹⁰ = [Zn²⁺] [CO₃²⁻]

We can find the moles of ZnBr₂ added = Moles of Zn²⁺ and moles of K₂CO₃ = Moles of CO₃²⁻ to find the moles of CO₃²⁻ that remains in solution, thus:

<em>Moles ZnB₂ (Molar mass: 225.2g/mol) = Moles Zn²⁺:</em>

6.63g ZnBr₂ * (1mol / 225.2g) = 0.02944moles Zn²⁺

<em>Moles K₂CO₃ = Moles CO₃²⁻:</em>

0.100L * (0.60mol/L) = 0.060 moles CO₃²⁻

Moles CO₃²⁻ in excess: 0.0600moles CO₃²⁻ - 0.02944moles =

0.03056moles CO₃²⁻ / 0.100L = 0.3056M = [CO₃²⁻]

Replacing in Ksp expression:

1.46x10⁻¹⁰ = [Zn²⁺] [0.3056M]

<h3>[Zn²⁺] = 4.78x10⁻¹⁰M</h3>

4 0
3 years ago
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