Choose the most alkaline substance in the group.

A pure copper penny contains approximately 2.9×1022 copper atoms. Use the following definitions to determine how many ______ of

hammer [34]

Complete question is;

A pure copper penny contains approximately 2.9 × 10^(22) copper atoms.

1 doz = 12

1 gross = 144

1 ream = 500

1 mol = 6.022 × 10^(23)

Use these definitions to determine the following:

A) How many dozens of copper atoms are in a penny.

B) How many gross of copper atoms are in a penny

C) How many reams of copper atoms are in a penny.

D) how many moles of copper atoms are in a penny?

All answers can be rounded to two significant figures

Answer:

A) 2.4 × 10^(21) dozens

B) 2.01 × 10^(20) gross

C) 5.8 × 10^(19) reams

D) 0.048 mol

Explanation:

A) A dozen contains 12.

Therefore, 2.9 × 10^(22) copper atoms will contain;

(2.9 × 10^(22))/12 dozens = 2.42 × 10^(21).

In 2 significant figures, we have;

2.4 × 10^(21) dozens

B) 1 gross = 144

Therefore, 2.9 × 10^(22) copper atoms will contain;

(2.9 × 10^(22))/144 gross ≈ 2.01 × 10^(20) gross

C) 1 ream = 500

Therefore, 2.9 × 10^(22) copper atoms will contain;

(2.9 × 10^(22))/500 reams = 5.8 × 10^(19) reams

D) 1 mol = 6.022 × 10^(23)

Therefore, 2.9 × 10^(22) copper atoms will contain;

(2.9 × 10^(22))/(6.022 × 10^(23)) = 0.048 mol

6 0

3 years ago

In a redox reaction, which particles are lost and gained in equal numbers

Anna35 [415]

In a redox reaction electrons are lost and gained in equal numbers. The species that is oxidized gives electrons to the species that is reduced. I hope this helps. Let me know if anything is unclear.

5 0

3 years ago

Find the molar enthalpy of formation for paraffin wax (C2H526)) given the following reaction

pogonyaev

Answer: The molar enthalpy of formation for paraffin wax is -2460.5 kJ

Explanation:

The balanced chemical reaction is,

$C_{25}H_{52}(g)+38O_2(g)\rightarrow 25CO_2(g)+26H_2O(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

$\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_{25}H_{52}}\times \Delta H_{C_{25}H_{52}})]$

where,

n = number of moles

$\Delta H_{O_2}=0$ (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

$-14800=[(25\times -393.5)+(26\times -285.5)]-[(38\times 0)+(1\times \Delta H_{C_{25}H_{52}})]$

$\Delta H_{C_{25}H_{52}}=-2460.5kJ/mol$

Therefore, the molar enthalpy of formation for paraffin wax is -2460.5 kJ

3 0

3 years ago

The gas molecules that make up air are in two rooms. The gas molecules in the kitchen are moving faster than the gas molecules i

Rainbow [258]

D, I think.
Because the bedroom is closer to the windows where is cold and wind went through, while the kitchen is a hot spots.

3 0

3 years ago

What happened to the volume of gas when the syringe was exposed to various temperature conditions? Using the concepts explored i

Delicious77 [7]

Answer:

Thus, when the volume of the gas is exposed to a temperature above -273.15 K, the volume increases linearly with the temperature.

Explanation:

The expression for Charles's Law is shown below:

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

This states that the volume of the gas is directly proportional to the absolute temperature keeping the pressure conditions and the moles of the gas constant.

Thus, when the volume of the gas is exposed to a temperature above -273.15 K, the volume increases linearly with the temperature.

For example , if the temperature of the gas is reduced to half, the volume also reduced to half.

At -273.15 K, according to Charles's law, it is possible to make the volume of an ideal gas = 0.

3 0

3 years ago