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3 years ago

An oxide of aluminum contains 0.545 g of Al and 0.485 g of O. Find the empirical formula

Chemistry

1 answer:

uysha [10]3 years ago

8 0

Answer:

The answer is: Al2O3

Explanation:

The data they give us is:

0.545 gr Al

0.485 gr O.

To find the empirical formula without knowing the grams of the compound, we find it per mole:

0.545 g Al * 1 mol Al / 27 g Al = 0.02 mol Al

0.485 g O * 1 mol O / 16 g O = 0.03 mol O

Then we must divide the results obtained by the lowest result, which in this case is 0.02:

0.02 mol Al / 0.02 = 1 Al

0.03 mol O / 0.02 = 1.5 O

Since both numbers have to give an integer, multiply by 2 until both remain integers:

1Al * 2 = 2Al

1.5O * 2 = 3O

Now the answer is given correctly:

Al2O3

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Titanium tribromide, titanium (III) bromide, or titanous bromide.

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3 years ago

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The standard free energy of formation, ΔG∘f, of a substance is the free energy change for the formation of one mole of the subst

OLEGan [10]

Answer:

B. 2 Na(s) + O₂(g) → Na₂O₂(s); ΔG∘f=−451.0 kJ/mol

D. 2 SO(g) + O₂(g) → 2 SO₂(g); ΔG°f=−600.4 kJ/mol

Explanation:

The spontaneity of a reaction is given by the value of the standard Gibbs free energy of the reaction (ΔG°rxn). The more negative is the ΔG°rxn, the more spontaneous is a reaction.

The ΔG°rxn can be calculated using the following expression:

ΔG°rxn = ∑np × ΔG°f(products) − ∑nr × ΔG°f(reactants)

By definition, the standard Gibbs free energy of formation of simple substances in their most stable state is zero. That is why, in the reaction of formation of a compound ΔG°rxn = ΔG°f(product).

Based on the standard free energies of formation, which of the following reactions represent a feasible way to synthesize the product?

A. N₂(g) + H₂(g) → N₂H₄(g); ΔG°f=159.3 kJ/mol.

Not feasible. ΔG°rxn = ΔG°f(product) > 0.

B. 2 Na(s) + O₂(g) → Na₂O₂(s); ΔG°f=−451.0 kJ/mol

Feasible. ΔG°rxn = ΔG°f(product) < 0.

C. 2 C(s) + 2 H₂(g) → C₂H₄(g); ΔG°f=68.20 kJ/mol

Not feasible. ΔG°rxn = ΔG°f(product) > 0.

D. 2 SO(g) + O₂(g) → 2 SO₂(g); ΔG°f=−600.4 kJ/mol

Feasible. ΔG°rxn = ΔG°f(product) < 0.

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4 years ago

Which of the following is a precipitation reaction?

erik [133]

Answer:

The precipitated are option a and d.

Explanation:

2 LiI(aq) +Hg2(NO3)2(aq) → Hg2I2(s) ↓ + 2 LiNO3(aq)

Cation Hg2+ 2 in the presence of iodide, a precipitated is formed.

Zn(s) + 2AgNO3(aq) → 2 Ag(s) ↓ +Zn(NO3)2(aq)

Zinc starts to get rid, and some white particles also stick to it. Afterwards the solution becomes cloudy and a precipitate appears, which is the solid silver

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3 years ago

The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of

gavmur [86]

Answer: The empirical formula for the given compound is $C_3H_6O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Conversion factor: 1 g = 1000 mg

Mass of $CO_2=6.32mg=0.00632g$

Mass of $H_2O=2.58g=0.00258g$

Mass of compound = 2.78 mg = 0.00278 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.00632 g of carbon dioxide, $\frac{12}{44}\times 0.00632=0.00172g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.00258 g of water, $\frac{2}{18}\times 0.00258=0.000286g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.00278) - (0.00172 + 0.000286) = 0.000774 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.00172g}{12g/mole}=1.43\times 10^{-4}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.000286g}{1g/mole}=2.86\times 10^{-4}moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.000774g}{16g/mole}=4.83\times 10^{-5}moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $4.83\times 10^{-5}mol$

For Carbon = $\frac{1.43\times 10^{-4}}{4.83\times 10^{-5}}=2.96\approx 3$

For Hydrogen = $\frac{2.86\times 10^{-4}}{4.83\times 10^{-5}}=5.92\approx 6$

For Oxygen = $\frac{4.83\times 10^{-5}}{4.83\times 10^{-5}}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is $C_3H_{6}O_1=C_3H_6O$

3 0

3 years ago

PLEASE SOMEONE HELP ME WITH THIS!

Bad White [126]

Answer:

1) 1.202 L , 2) 1.291 dg , 3) 204.877 and 4) 1.04x $10^{3\\}$

Explanation:

You need to review about conversion factors and how to use them in the correct order. You can cancel the units and get the ones that you need if you use the appropriate conversion factors, remember is a number that you can use to multiply or divide.

For your exercise:

1) The conversion factor is: 1 L = 1000 mL

You will need to divide by 1000 mL to obtain liters L

1202.57120 mL x $\frac{1 L}{1000 mL}$ = 1.202 L

2) The conversion factor is: 1 g = 10 dg

0.1290743 g x $\frac{10 dg}{ 1 g}$ = 1.291 dg

For the next exercises, you need to follow some rules:

1. All numbers that are different from Zero (non-zero digits) are significant figures.

2.The Zeros between non-zeros digits (Imbedded zeros) always are significant, 2007.

3. If you want to be specific and want some zeros to be significant you need to add a decimal point. For example 500. or 500.0

4. Leading zeros (to the left) are not significant.

5. Trailing zeros (zeros to the right) in a whole number without decimal point are not significant.

3) 843.062 - 638.1848 = 204.8772

Now if we round to 6 significant figures we get 204.877

4)123.0 x 8.43 = 1036.89

Now we round to 3 significant figures because 8.43 has the least significant figures.

1.04x $10^{3}$

4 0

3 years ago