All categories

3 years ago

A sample of aluminum foil contains 7.90 x 10^23 atoms. What is the mass of the foil?

Chemistry

1 answer:

Zanzabum3 years ago

4 0

Answer:

35.3 g

Explanation:

Step 1: Given data

Number of atoms of aluminum: 7.90 × 10²³ atoms

Step 2: Calculate the number of moles corresponding to 7.90 × 10²³ atoms of aluminum

We will use Avogadro's number: there are 6.02 × 10²³ atoms of aluminum in 1 mole of atoms of aluminum.

7.90 × 10²³ atoms × (1 mol/6.02 × 10²³ atoms) = 1.31 mol

Step 3: Calculate the mass corresponding to 1.31 moles of aluminum

The molar mass of Al is 26.98 g/mol.

1.31 mol × 26.98 g/mol = 35.3 g

You might be interested in

A piece of balsa wood has a mass of 15.196 g and a volume of 0.1266 L. What is the density in g/mL?

Nataliya [291]

Density = mass / volume

Since the answer must be in g/ml
Convert volume to ml
0.1266 x 1000 = 126.6 ml

15.196 / 126.6

= 0.126 g/ml

4 0

3 years ago

The number of protons in an atom is defined by it's

swat32

The atomic number defines the number of protons in an atom.

6 0

3 years ago

217 mL of neon gas at 551 celsius is brought to standard temperature (0 celsius) while the pressure is held constant. What is th

pantera1 [17]

Answer: The new volume is 72 ml

Explanation:

To calculate the final volume of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1=217ml\\T_1=551^oC=(551+273)K=824K\\V_2=?\\T_2=0^0C=(0+273)K=273K$

Putting values in above equation, we get:

$\frac{217ml}{824K}=\frac{V_2}{273K}\\\\V_2=72ml$

Thus the new volume is 72 ml

7 0

2 years ago

Consider the electrolysis of molten barium chloride, BaCl2. (a) Write the half-reactions. (b) How many grams of barium metal can

Zielflug [23.3K]

Answer: a) Cathode(-): $Ba^+^2+2e^-\rightarrow Ba$

Anode(+): $2Cl^-\rightarrow Cl_2+2e^-$

b) 0.640 grams of Ba will be deposited.

Explanation: a) The problem is based on Faraday law of electrolysis. Molten barium chloride has $Ba^+^2$ ion and $Cl^-$ ion. Barium ion is reduced and chloride is oxidized. Reduction takes place at cathode and oxidation at anode. So, the half reactions will be:

Cathode(-): $Ba^+^2+2e^-\rightarrow Ba$

Anode(+): $2Cl^-\rightarrow Cl_2+2e^-$

b) The question asks, how many grams of barium metal can be produced by supplying 0.50 ampere for 30 minutes.

From the Cathode half-reaction, 1 mole of Ba is deposited by 2 moles of electrons and we know that 1 mole of electron carries one Faraday that is 96485 Coulomb.

Coulombs for 2 moles of electrons will be = 2*96485 C = 192970 C

So, we can say that, 192970 C will deposit 1 mole of Ba metal.

Total available coulombs can be calculated using the formula:

q=i*t

where, q is electric charge in coulomb, i is current in ampere and t is time in seconds.

$q=q=0.50A*30min(\frac{60sec}{1min})$