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2 years ago

Calculate the number of oxygen atoms in a 80.0 g sample of scheelite (CaWo).

Chemistry

1 answer:

castortr0y [4]2 years ago

3 0

Answer:

Explanation:

scheelite is CaWO₄

Mol weight = 288

80 g of scheelite = 80 / 288 = 27.77 x 10⁻² moles

27.77 x 10⁻² moles of scheelite = 27.77 x 10⁻² x 6.02 x 10²³ molecules of scheelite

= 167.17 x 10²¹ molecules of scheelite

1 molecule of scheelite contains 4 atoms of oxygen

167.77 x 10²¹ molecules of scheelite contains 4 x 167.77 x 10²¹ atoms of oxygen .

= 671.08 x 10²¹ atoms of oxygen .

= 671 x 10²¹ atoms .

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2 years ago

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3 years ago

Read 2 more answers

How many mL of a 1.48 M calcium hydroxide solution are needed to neutralize 36.0 mL of a 1.63 M hydrochloric acid solution

Lelechka [254]

The volume (in mL) of calcium hydroxide, Ca(OH)₂ needed for the reaction is 19.8 mL

<h3>Balanced equation </h3>

2HCl + Ca(OH)₂ —> CaCl₂ + 2H₂O

From the balanced equation above,

The mole ratio of the acid, HCl (nA) = 2
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<h3>How to determine the volume of Ca(OH)₂ </h3>

Molarity of base, Ca(OH)₂ (Mb) = 1.48 M
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Volume of base, Ca(OH)₂ (Vb) =?

MaVa / MbVb = nA / nB

(1.63 × 36) / (1.48 × Vb) = 2

58.68 / (1.48 × Vb) = 2

Cross multiply

2 × 1.48 × Vb = 58.68

2.96 × Vb = 58.68

Divide both side by 2.96

Vb = 58.68 / 2.96

Vb = 19.8 mL

Learn more about titration:

brainly.com/question/14356286

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