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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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According to valence bond theory, which orbitals overlap in the formation of the bond in HCl?
SIZIF [17.4K]
A) 1s on H and 3p on Cl

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What volume of oxygen gas is released at STP if 10.0 g of potassium chlorate is decomposed? (The molar mass of KClO3 is 122.55 g
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<span>KCl<span>O3</span><span>(s)</span>+Δ→KCl<span>(s)</span>+<span>32</span><span>O2</span><span>(g)</span></span>

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Explanation:

We assume that the reaction as written proceeds quantitatively.

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And thus <span><span>32</span>×0.0816⋅mol</span> dioxygen are produced, i.e. <span>0.122⋅mol</span>.

At STP, an Ideal Gas occupies a volume of <span>22.4⋅L⋅mo<span>l<span>−1</span></span></span>.

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8 0
3 years ago
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