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2 years ago

A compound is found to contain 43.66% of P and 56.33% of O.

Chemistry

1 answer:

jek_recluse [69]2 years ago

6 0

Answer: P2O5 is the empirical formula.

Explanation: When given percentages you can assume that many grams of each atom are in the compound. Then you divide grams by the molar mass of each element, giving you moles. Once you have moles, divide by the smaller molar amount, which should give you 1 mol of Phosphorus and 2.5 mol of Oxygen. Then multiply by 2 in order for both moles to be a whole number. This gets you 2 and 5.

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alexandr402 [8]

Answer:

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6 0

2 years ago

Calculate the number of formula units in 2.50 mol NaNO

marta [7]

The number of formula units in 2.50 mol of the compound is 15.1 * 10^23.

The question is unclear whether NaNO2 or NaNO3 is implied. However,in either case, the solution applies equally.

6.02 * 10^23 formula units of the compound are contained in 1 mole

x formula units are contained in 2.5 moles of the compound

x = 6.02 * 10^23 formula units * 2.5 moles/ 1 mole

x = 15.1 * 10^23 formula units of the compound.

Learn more; brainly.com/question/9743981

3 0

2 years ago

Which of the following increases as you move up a column of the periodic table?

luda_lava [24]

Answer:

electronegativity increases

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2 years ago

DISCUSSION QUESTION

andrey2020 [161]

No, I don’t think so. My reason being that some of these viruses are not harmful, an example is bacteriophage which kills bacteria.

6 0

3 years ago

What quantity of sodium azide in grams is required to fill a 56.0 liters air bag with nitrogen gas at 1.00 atm and exactly 0 °C:

Margarita [4]

Answer:

108.6 g

Explanation:

2NaN₃(s) → 2Na(s) + 3N₂(g)

First we use the PV=nRT formula to calculate the number of nitrogen moles:

P = 1.00 atm

V = 56.0 L

n = ?

R = 0.082 atm·L·mol⁻¹·K⁻¹

T = 0 °C ⇒ 0 + 273.2 = 273.2 K

Inputting the data:

1.00 atm * 56.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.2 K

n = 2.5 mol

Then we convert 2.5 moles of N₂ into moles of NaN₃, using the stoichiometric coefficients of the balanced reaction:

2.5 mol N₂ * $\frac{2molNaN_3}{3molN_2}$ = 1.67 mol NaN₃

Finally we convert 1.67 moles of NaN₃ into grams, using its molar mass:

1.67 mol * 65 g/mol = 108.6 g

6 0

2 years ago