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3 years ago

A percent composition analysis yields 46.7% nitrogen and 53.3% oxygen. What is the empirical formula for the compound

1 answer:

6 0

Empirical formula is the simplest ratio of whole numbers of components in a compound
calculating for 100 g of compound
N O
mass 46.7 g 53.3 g
number of moles 46.7 g / 14 g/mol 53.3 g / 16 g/mol
= 3.33 = 3.33
divide by the least number of moles
3.33/ 3.33 = 1.00 3.33/3.33 = 0
number of atoms
N - 1
O - 1
therefore empirical formula is NO

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3 years ago

Helppppp pleaseeee xxxxxx

kogti [31]

Answer:

2812.6 g of H₂SO₄

Explanation:

From the question given above, the following data were obtained:

Mole of H₂SO₄ = 28.7 moles

Mass of H₂SO₄ =?

Next, we shall determine the molar mass of H₂SO₄. This can be obtained as follow:

Molar mass of H₂SO₄ = (1×2) + 32 + (16×4)

= 2 + 32 + 64

= 98 g/mol

Finally, we shall determine the mass of H₂SO₄. This can be obtained as follow:

Mole of H₂SO₄ = 28.7 moles

Molar mass of H₂SO₄ =

Mass of H₂SO₄ =?

Mole = mass / Molar mass

28.7 = Mass of H₂SO₄ / 98

Cross multiply

Mass of H₂SO₄ = 28.7 × 98

Mass of H₂SO₄ = 2812.6 g

Thus, 28.7 mole of H₂SO₄ is equivalent to 2812.6 g of H₂SO₄

3 0

3 years ago

A plane takes 230 passengers from New York to Chicago in 3 hours.

devlian [24]

Answer:263mph

Explanation:

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2 years ago

If an object has a density of 0.00018 g/cm cubed, what density would it be if it had a mass of 10kg

Vladimir79 [104]

Answer:

Explanation:

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2 years ago

Extend your thinking: The slow decay of radioactive materials can be used to find the age of rocks, fossils, and archaeological

lisov135 [29]

The age in years of the Egyptian papyrus, the Aboriginal charcoal, the Mayan headdress, and the Neanderthal skull are 4000 years, 13106.5years , 2040 years, and 30353 years respectively.

<h3>What is the half-life of a radioactive material?</h3>

The half-life of a radioactive material is the time taken for half the atoms present in the material to decay or disintegrate.

The half-life, $t_\frac{1}{2}$ , the age, t, and amount remaining, $A_{r}$ , of a radioactive material are related by the formula below:

$t = \frac{t_\frac{1}{2}*A_{r}}{-ln2}$

Half-life of carbon-14 = 6000 years

For the Egyptian papyrus with 63% of its original carbon-14 atoms: $t = \frac{6000*0.63}{-0.63} = 4000\:years$

For the Aboriginal charcoal with 22% of its original carbon-14 atoms:

$t = \frac{6000*0.22}{-0.63} = 13106.5\:years$

For the Mayan headdress with 79% of its original carbon-14 atoms:

$t = \frac{6000*0.79}{-0.63} = 2040\:years$

Neanderthal skull with 3% of its original carbon-14 atoms:

$t = \frac{6000*0.03}{-0.63} = 30353\:years$

Therefore, the age in years of the Egyptian papyrus, the Aboriginal charcoal, the Mayan headdress, and the Neanderthal skull are 4000 years, 13106.5years , 2040 years, and 30353 years respectively.

Learn more about half-life at: brainly.com/question/26689704

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2 years ago