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2 years ago

12

A 10.50 gram sample of a compound is decomposed to yield 3.40 g Na, 2.37 g S, and 4.73 g O. What is the mass percentage of each

element in this compound?

1 answer:

9966 [12]2 years ago

6 0

Answer:

Na = 32.4% , % S = 22.6% and %O = 45.0%

Explanation:

% Na = 3.4/10.5. × 100%

= 32.4%

%S = 2.37/10.5 × 100%

= 22.6%

% O= 4.73/10.5 × 100%

= 45.0%xplanation:

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The amount of which subatomic particle is different between an atom and its ion?

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3 years ago

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andrew11 [14]

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3 years ago

Bob measured out 1.60 grams of sodium. He calculates that 1.60 g of

saul85 [17]

Answer:

84.8%

Explanation:

Step 1: Given data

Bob measured out 1.60 g of Na. He forms NaCl according to the following equation.

Na + 1/2 Cl₂ ⇒ NaCl

According to this equation, he calculates that 1.60 g of sodium should produce 4.07 g of NaCl, which is the theoretical yield. However, he carries out the experiment and only makes 3.45 g of NaCl, which is the real yield.

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3 years ago

Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g)

Fynjy0 [20]

Answer:

Rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

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According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)

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Given that the rate of disappearance of oxygen = $-\frac{d[O_{2} ]}{dt}$ = 3.64 x 10⁻³ M/s

So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = ?

from equation (1) we can write

$\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]$

⇒ $\frac{d[SO_{3}] }{dt}$ = 2 x 3.64 x 10⁻³ M/s

⇒ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

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3 years ago