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andrew11 [14]
4 years ago
8

A new penny has a mass of 2.49 grams and occupies 0.349 cm cubed. If pure copper has a density of 8.96 g/cm cubed, is the new pe

nny put copper?
Chemistry
1 answer:
lys-0071 [83]4 years ago
8 0

Answer: The new penny is not pure copper

Explanation:

Density \rho is defined as a relation between the mass m and the volume V:

\rho=\frac{m}{V}

Now, we are told the density of pure copper is:

\rho_{copper}=8.96 g/cm^{3}

And we are given the mass and volume of the new penny, with which we can calculate its density:

\rho_{penny}=\frac{m_{penny}}{V_{penny}}=\frac{2.49 g}{0.349 cm^{3}}

\rho_{penny}=7.13 g/cm^{3} As we can see the density of this penny is not equal to the density of pure copper, hence the new penny is not pure copper.

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A solution of NaOH is titrated with H2SO4. It is found that 20.05 mL of 0.3564 M H2SO4 solution is equivalent to 43.42 mL of NaO
Darya [45]

Answer : The concentration of NaOH is, 0.336 M

Explanation:

To calculate the concentration of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=0.3564M\\V_1=20.05mL\\n_2=1\\M_2=?\\V_2=43.42mL

Putting values in above equation, we get:

2\times 0.3564M\times 20.05mL=1\times M_2\times 43.42mL

M_2=0.336M

Thus, the concentration of NaOH is, 0.336 M

3 0
4 years ago
In the Mond process for the purification of nickel, carbon monoxide is reacted with heated nickel to produce Ni(CO)4, which is a
igor_vitrenko [27]

<u>Answer:</u> The equilibrium constant for this reaction is 1.068\times 10^{6}

<u>Explanation:</u>

The equation used to calculate standard Gibbs free change is of a reaction is:

\Delta G^o_{rxn}=\sum [n\times \Delta G^o_{(product)}]-\sum [n\times \Delta G^o_{(reactant)}]

For the given chemical reaction:

Ni(s)+4CO(g)\rightleftharpoons Ni(CO)_4(g)

The equation for the standard Gibbs free change of the above reaction is:

\Delta G^o_{rxn}=[(1\times \Delta G^o_{(Ni(CO)_4(g))})]-[(1\times \Delta G^o_{(Ni(s))})+(4\times \Delta G^o_{(CO(g))})]

We are given:

\Delta G^o_{(Ni(CO)_4(g))}=-587.4kJ/mol\\\Delta G^o_{(Ni(s))}=0kJ/mol\\\Delta G^o_{(CO(g))}=-137.3kJ/mol

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(1\times (-587.4))]-[(1\times (0))+(4\times (-137.3))]\\\\\Delta G^o_{rxn}=-38.2kJ/mol

To calculate the equilibrium constant (at 58°C) for given value of Gibbs free energy, we use the relation:

\Delta G^o=-RT\ln K_{eq}

where,

\Delta G^o = Standard Gibbs free energy = -38.2 kJ/mol = -38200 J/mol  (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = 8.314 J/K mol

T = temperature = 58^oC=[273+58]K=331K

K_{eq} = equilibrium constant at 58°C = ?

Putting values in above equation, we get:

-38200J/mol=-(8.314J/Kmol)\times 331K\times \ln K_{eq}\\\\K_{eq}=e^{13.881}=1.068\times 10^{6}

Hence, the equilibrium constant for this reaction is 1.068\times 10^{6}

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3 years ago
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Otrada [13]

Answer:

A. They are malleable, ductile, and conduct electricity.

Explanation:

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Answer: In order for reflection to occur, the wave must “bounce off” an object

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