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3 years ago

What is the molar mass of aluminum (Al)

Chemistry

2 answers:

Nookie1986 [14]3 years ago

8 0

Answer:

2.68573957

Explanation:

Papessa [141]3 years ago

6 0

Answer:

26.981539 u

Explanation:

One mole of Al atoms has a mass in grams that is numerically equivalent to the atomic mass of aluminum. The periodic table shows that the atomic mass (rounded to two decimal points) of Al is 26.98, so 1 mol of Al atoms has a mass of 26.98 g.

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How many electrons in an atom could have these sets of quantum numbers? n=2?

slavikrds [6]

A maximum of 8 electrons can share the quantum number n = 2.

Principal Quantum number has a symbol of "n". It tells you the energy level on which an electron resides. You need to determine exactly how many orbitals you have in this energy level before you can determine the number of electrons that can share the value of n.

The number of orbitals you get per energy level can be found using this formula:

no. of orbitals=n²

Each orbital can hold a maximum of two electrons, the formula would be:

no. of electrons=2n²

Using the given formulas:

no. of orbitals = n² = 2² = 4

no. of electrons =2 * 4 = 8

6 0

4 years ago

suppose a 22.092 g sample of 1 1 mixture of acetylferrocene and ferrocene was sepeerated by column chromatography and the recove

maksim [4K]

Answer:

Percentage recovery of acetylferrocene = 81.6%

Explanation:

Mass of the sample mixture = 22.092 g

Ratio of mixture of acetylferrocene and ferrocene = 1 : 1

This means that the sample conatains equal amounts of acetylferrocene and ferrocene.

Therefore the mass of each sample in the mixture = 22.092 g / 2 = 11.046 g

Mass of acetylferrocene recovered = 9.017 g

Percentage recovery of acetylferrocene = (mass of recovered/ mass in sample) * 100%

Percentage recovery of acetylferrocene = (9.017 g / 11.046 g) *100%

Percentage recovery of acetylferrocene = 81.6%

3 0

4 years ago

How many moles of ammonia are in 735. mL of a 4.25 M aqueous ammonia solution?

zysi [14]

Answer:

*moles = molarity(M=moles/L) * volume (L)

so, you have 4.25 * 0.735 moles

*NH3

* M=n/v

*To determine the number of significant figures in a number use the following 3 rules: Non-zero digits are always significant. Any zeros between two significant digits are significant. A final zero or trailing zeros in the decimal portion ONLY are significant.

6 0

3 years ago

When a base is dissolved in water, what is released? A. H+ B. OH− C. CO2

Svet_ta [14]

Answer:Your answer is B

Hope this helps have a great day<3

Explanation:

8 0

3 years ago

Read 2 more answers

Ethylene (CH2CH2) is the starting point for a wide array of industrial chemical syntheses. For example, worldwide about 8.0 x 10

mylen [45]

Answer: The percent by mass of ethylene in the equilibrium gas mixture is 3.76 %

Explanation:

We are given:

Initial partial pressure or ethane = 24.0 atm

The chemical equation for the dehydration of ethane follows:

$C_2H_6(g)\rightleftharpoons C_2H_4(g)+H_2(g)$

Initial: 24.0

At eqllm: 24-x x x

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{C_2H_4}\times p_{H_2}}{p_{C_2H_6}}$

We are given:

$K_p=0.040$

Putting values in above expression, we get:

$0.040=\frac{x\times x}{24-x}\\\\x^2+0.04x-0.96=0\\\\x=0.96,-1$

Neglecting the value of x = -1 because partial pressure cannot be negative.

So, partial pressure of hydrogen gas at equilibrium = x = 0.96 atm

Partial pressure of ethylene gas at equilibrium = x = 0.96 atm

Partial pressure of ethane gas at equilibrium = (24-x) = (24 - 0.96) atm = 23.04 atm

To calculate the number of moles, we use the equation given by ideal gas, which follows:

$PV=nRT$ .........(1)

To calculate the mass of a substance, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ..........(2)

For ethane gas:

We are given:

$P=23.04atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$23.04atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{23.04\times 30.0}{0.0821\times 1073}=7.85mol$

We know that:

Molar mass of ethane gas = 30 g/mol

Putting values in equation 2, we get:

$7.85mol=\frac{\text{Mass of ethane gas}}{30g/mol}\\\\\text{Mass of ethane gas}=(7.85mol\times 30g/mol)=235.5g$

For ethylene gas:

We are given:

$P=0.96atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$0.96atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{0.96\times 30.0}{0.0821\times 1073}=0.33mol$

We know that:

Molar mass of ethylene gas = 28 g/mol

Putting values in equation 2, we get:

$0.33mol=\frac{\text{Mass of ethylene gas}}{28g/mol}\\\\\text{Mass of ethylene gas}=(0.33mol\times 28g/mol)=9.24g$

For hydrogen gas:

We are given:

$P=0.96atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$0.96atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{0.96\times 30.0}{0.0821\times 1073}=0.33mol$

We know that:

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 2, we get:

$0.33mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=(0.33mol\times 2g/mol)=0.66g$

To calculate the mass percentage of ethylene in equilibrium gas mixture, we use the equation:

$\text{Mass percent of ethylene gas}=\frac{\text{Mass of ethylene gas}}{\text{Mass of equilibrium gas mixture}}\times 100$

Mass of equilibrium gas mixture = [235.5 + 9.24 + 0.66] = 245.4 g

Mass of ethylene gas = 9.24 g

Putting values in above equation, we get:

$\text{Mass percent of ethylene gas}=\frac{9.24g}{245.5g}\times 100=3.76\%$

Hence, the percent by mass of ethylene in the equilibrium gas mixture is 3.76 %

5 0

3 years ago