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qaws [65]
2 years ago
6

7. How many moles of mercury(ii) oxide, hgo, are needed to produce 12. 5 g of oxygen, o2? 2 hgo(s) --->2 hg(l) o2(g)

Chemistry
1 answer:
svetoff [14.1K]2 years ago
5 0

Moles are the division of the mass and the molar mass. The moles of mercury (ii) oxide in the decomposition reaction needed to produce oxygen are 0.781 moles.

<h3>What is a decomposition reaction?</h3>

A decomposition reaction is a breakdown of the reactant into simpler products. The decomposition of mercury (ii) oxide can be shown as:

2HgO(s) → 2Hg(l) + O₂(g)

From the reaction, it can be said that 2 moles of mercury (ii) oxide decomposes to produce 1 mole of oxygen.

The moles of oxygen that needs to be produced are calculated as:

Moles = mass ÷ molar mass

= 12.5 gm ÷ 32 gm/mol

= 0.39 moles

0.39 moles of oxygen are needed to be produced.

From the stoichiometric coefficient of the reaction, the moles of HgO is calculated as: 2 × 0.39 = 0.781 moles

Therefore, 0.781 moles of HgO are required in the reaction.

Learn more about moles here:

brainly.com/question/3801333

#SPJ4

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The________ cation will precipitate after the_________ step. Then the _________cation will precipitate after the________ step.
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f. Sn^4+

c. second

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8 0
3 years ago
Convert each quantity to the indicated units. a. 3.01g to cg. b. 6200m to km
miss Akunina [59]

a. 301 cg

b. 6.2 km

Explanation:

a. knowing that 1 gram (g) is equal to 100 centigrams (cg) we devise the following reasoning:

if        1 g is equal to 100 cg

then  3.01 g are equal to X cg

X = (3.01 × 100) / 1 = 301 cg

b. knowing that 1 kilometer (km) is equal to 1000 meters (m) we devise the following reasoning:

if         1 km is equal to 1000 m

then   Y km are equal to 6200 m

Y = (6200 × 1) / 1000 = 6.2 km

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Help please please please please
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Answer:

D. 18.7 grams

Explanation:

The coefficients are the key.

Create a proportion with them, and the molar mass, and then solve for x:

\frac{2}{3}  =  \frac{x}{28}  \\ 3x = 2(28) \\ 3x = 56 \\ x = 18.666... = 18.7

7 0
4 years ago
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