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aleksklad [387]
2 years ago
10

Will has been told to check the oxygen level in a confined space, before workers are allowed to enter. Will should verify that t

he space has at least ____ % oxygen by volume.
Chemistry
1 answer:
loris [4]2 years ago
8 0
19.5%.

If you need a full explanation, let me know :]
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Colt1911 [192]
Rise, decrease, away from ocean, towards land
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Which is not a descriptor or tactic used by abusers?
attashe74 [19]
It should be B. All of the other choices are used often by abusers
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3 years ago
A compound is found to contain 50. 05% sulfur and 49. 95% oxygen by mass. What is the empirical formula for this compound? SO S2
jekas [21]

The empirical formula for the given compound has been \rm SO_2. Thus, option C is correct.

The empirical formula has been the whole unit ratio of the elements in the formula unit.

<h3>Computation for the Empirical formula</h3>

The given mass of Sulfur has been, 50.05 g

The given mass of oxygen has been 49.95 g.

The moles of elements in the sample has been given by:

\rm Moles=\dfrac{Mass}{Molar\;mass}

  • Moles of Sulfur:

\rm Moles\;S=\dfrac{50.05}{32}\\&#10; Moles\;S=1.56\;mol

The moles of sulfur in the unit has been 1.56 mol.

  • Moles of Oxygen:

\rm Moles\;O=\dfrac{49.95}{16} \\&#10;Moles\;O=3.12\;mol

The moles of oxygen in the unit has been 3.12 mol.

The empirical formula unit has been given as:

\rm S_{1.56}O_{3.12}=SO_2

Thus, the empirical formula for the given compound has been \rm SO_2. Thus, option C is correct.

Learn more about empirical formula, here:

brainly.com/question/11588623

8 0
2 years ago
aspirin C6H4(CO2)(CO2CH3),can be prepared in the chemistry laboratory by the reactions of salicylic acid, C6H4(CO2H)(OH),with ac
Ket [755]
262mol 1=kg
g=1000
from stoichoimetry

x=102*1000/360
x=102000/360
x=283.33

density =m/v
=283.33/1.082
=262mol

8 0
3 years ago
The percent composition of calcium is ?
DochEvi [55]

Answer: 40.1%

Explanation: The mass of calcium in this compound is equal to 40.1 grams because there's one atom of calcium present and calcium has an atomic mass of 40.1 . The molar mass of the compound is 100.1 grams. Using the handy equation above, we get: Mass percent = 40.1 g Ca⁄100.1 g CaCO3 × 100% = 40.1% Ca.

5 0
3 years ago
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