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erik [133]
3 years ago
6

If 11.87 grams of sand (SiO2) were contained in the mixture, how many atoms of oxygen were in the mixture?

Chemistry
1 answer:
ozzi3 years ago
6 0

Answer:

2.38x10²³ atoms oxygen

Explanation:

To solve this question we need to convert the mass of sand to moles using its molar mass (Molar mass SiO₂ = 60.08g/mol). Twice these moles are the moles of Oxygen and using Avogadro's number we can find the amount of atoms of Oxygen in the mixture:

<em>Moles SiO₂:</em>

11.87g * (1mol / 60.08g) = 0.1976moles SiO₂

<em>Moles Oxygen:</em>

0.1976moles SiO₂* 2 = 0.3951moles oxygen

<em>Atoms oxygen:</em>

0.3951moles oxygen * (6.022x10²³atoms / 1mol O₂) =

<h3>2.38x10²³ atoms oxygen</h3>
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Vlad [161]

Answer:

52 amu

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[89/100 * 52] + [8/100 * 49] + [3/100 * 50]

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Answer:

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Calculate the mass of MNO2 needed to produce 25.0 g of Cl2
Vedmedyk [2.9K]
I will solve this question assuming the reaction equation look like this:
<span>MnO2 + 4 HCl ---> MnCl2 + Cl2 + 2 H2O. 
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Now, we use the equation:
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