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2 years ago

The mass of a solid copper penny is 3.46 g. how many copper atoms are in the penny?

Chemistry

2 answers:

solong [7]2 years ago

3 0

3.46 g Cu 1 mol Cu 6.02*10^23 atoms Cu =
63.55 g Cu 1 mol Cu

3.28*10^22 atoms Cu

garri49 [273]2 years ago

3 0

Answer:

n° atoms = 3.275x10²² atoms

Explanation:

To do this, we need to use the molecular weight of the copper, which is 63.55 g/mol.

Now, to know the number of atoms of any compound, we use the avogadro number which is relationed to the number of moles of any compound so:

1 mol = 6.02x10²³ atoms

In order to know the atoms, we need to calculate the moles of copper with the following expression:

n = m/MW

Doing so, we have the moles, and then, use the avogadro number to calculate the atoms:

First the moles:

n = 3.46/63.55 = 0.0544 moles

Now, let's calculate the atoms:

n° atoms = n * Av. number

n° atoms = 0.0544 ¨* 6.02x10²³

n° atoms = 3.275x10²² atoms

This would be the atoms in the penny.

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If this decay has a half life of 2.60 years, what mass of 72.5 g of sodium 22 will remain after 15.6 years

Vlad1618 [11]

Sodium-22 remain : 1.13 g

<h3>Further explanation </h3>

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.

Usually, radioactive elements have an unstable atomic nucleus.

General formulas used in decay:

$\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{T/t\frac{1}{2} }}}$

T = duration of decay

t 1/2 = half-life

N₀ = the number of initial radioactive atoms

Nt = the number of radioactive atoms left after decaying during T time

half-life = t 1/2=2.6 years

T=15.6 years

No=72.5 g

$\tt Nt=72.5.\dfrac{1}{2}^{15.6/2.6}\\\\Nt=72.5.\dfrac{1}{2}^6\\\\Nt=1.13~g$

8 0

2 years ago

Consider the following reactions and their respective equilibrium constants: NO(g)+12Br2(g)⇌NOBr(g)Kp=5.3 2NO(g)⇌N2(g)+O2(g)Kp=2

maxonik [38]

Answer:

Equilibrium constant of the given reaction is $1.3\times 10^{-29}$

Explanation:

$NO+\frac{1}{2}Br_{2}\rightleftharpoons NOBr$ .... $(K_{p})_{1}=5.3$

$2NO\rightleftharpoons N_{2}+O_{2}$ .... $(K_{p})_{2}=2.1\times 10^{30}$

The given reaction can be written as summation of the following reaction-

$2NO+Br_{2}\rightleftharpoons 2NOBr$

$N_{2}+O_{2}\rightleftharpoons 2NO$

......................................................................................

$N_{2}+O_{2}+Br_{2}\rightleftharpoons 2NOBr$

Equilibrium constant of this reaction is given as-

$\frac{[NOBr]^{2}}{[N_{2}][O_{2}][Br_{2}]}$

$=(\frac{[NOBr]}{[NO][Br_{2}]^{\frac{1}{2}}})^{2}(\frac{[NO]^{2}}{[N_{2}][O_{2}]})$

$=\frac{(K_{p})_{1}^{2}}{(K_{p})_{2}}$

$=\frac{(5.3)^{2}}{2.1\times 10^{30}}=1.3\times 10^{-29}$

7 0

2 years ago

Read 2 more answers

A 50.0 mL sample of 0.600 M calcium hydroxide is mixed with 50.0 mL sample of 0.600 M hydrobromic acid in a Styrofoam cup. The t

TEA [102]

Explanation:

The reaction equation will be as follows.

$Ca(OH)_{2}(aq) + 2HBr(aq) \rightarrow CaBr_{2}(aq) + 2H_{2}O(l)$

So, according to this equation, 1 mole $Ca(OH)_{2}$ = 2 mol HBr = 1 mol $CaBr_{2}$

Therefore, calculate the number of moles of calcium hydroxide as follows.

No. of moles of $Ca(OH)_{2}$ = $V \times Molarity$

= $50 \times 0.6$

= 30 mmol

Similarly, calculate the number of moles of HBr as follows.

No. of moles of HBr = $M \times V$

= $50 \times 0.6$

= 30 mmol

This means that the limiting reactant is HBr.

So, no. of moles of $CaBr_{2}$ = $30 \times \frac{1}{2}$

= 15 mmol

Hence, calculate the amount of heat released as follows.

Heat released in the reaction(q) = $m \times s \times \Delta T$

as, m = mass of solution

and, Density = $\frac{mass}{volume}$

or, mass = Density × Volume

= 1.08 g/ml \times (50 + 50) ml

= 108 g

where, s = specific heat of solution = 4.18 j/g.k

and, change in temperature $\Delta T$ = $(26 - 23)^{o}C$

= $3
^{o}C$

Hence, the heat released will be as follows.

q = $m \times s \times \Delta T$

q = $108 \times 4.18 \times 3^{o}C$

= 1354.32 joule

or, = 1.354 kJ (as 1 kJ = 1000 J)

Also, $\Delta H_{rxn}$ = $\frac{-q}{n}$

= $\frac{-1.354}{15 \times 10^{-3}}$

= -90.267 kJ/mol

Thus, we can conclude that the enthalpy change for the given reaction is -90.267 kJ/mol.

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2 years ago

Compare and contrast steak and juice

blsea [12.9K]

Steak is a meat and juice is a liquid

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2 years ago

84. Heavy water, D2O (molar mass = 20.03 g mol–1), can be separated from ordinary water, H2O (molar mass = 18.01), as a result o

ipn [44]

Answer:

relative rate of diffusion is 1.05

Explanation:

According to Graham's law of difussion:

Rate of diffusion is inversely proportional to the square root of molecular weight of a molecule.

For two given molecules:

$\frac{(rate)_{1}}{(rate)_{2}}=\frac{\sqrt{M_{2}} }{\sqrt{M_{1}}}$

The given molecules are

Water = 18.01

Heavy water =20.03

Thus the relative rate of diffusion will be:

$\frac{(rate)_{water}}{(rate)_{heavywater}}=\sqrt{ \frac{20.03}{18.01}}=1.05$

8 0

2 years ago