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3 years ago

Suggest 2 reasons why copper should not be disposed in landfill sites

Chemistry

2 answers:

lidiya [134]3 years ago

8 0

It's a reliable source to many of products and it's value is far to hight to simply abandoned

SIZIF [17.4K]3 years ago

6 0

Because it's a waste. And it's medal instead of disposing just reuse it and also it can affect people Like high exposure of copper can be dangerous

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The enthalpy change for converting 1.00 mol of ice at -50.0 ∘c to water at 60.0∘c is ________ kj. the specific heats of ice, wat

guajiro [1.7K]

First, we have to get:

1- The heat required to increase T of ice from -50 to 0 °C:

according to q formula:

q1 = m*C*ΔT

when m is the mass of ice = mol * molar mass

= 1 mol * 18 mol/g

= 18 g

and C is the specific heat capacity of ice = 2.09 J/g-K

and ΔT change in temperature = 0- (-50) = 50°C

by substitution:

∴q1 = 18 g * 2.09 J/g-K *50°C

= 1881 J = 1.881 KJ

2- the heat required to melt this mass of ice is :

q2 = n*ΔHfus

when n is the number of moles of ice = 1 mol

and ΔHfus = 6.01 KJ/mol

by substitution:

q2 = 1 mol * 6.01 KJ/mol

= 6.01 KJ

3- the heat required to increase the water temperature from 0°C to 60 °C is:

q3 = m*C*ΔT

when m is the mass of water = 18 g

C is the specific heat capacity of water = 4.18 J/g-K

ΔT is the change of Temperature of water = 60°C - 0°C = 60°C

by substitution:

∴q3 = 18 g * 4.18 J/g-K * 60°C

= 4514 J = 4.514 KJ

∴the total change of enthalpy = q1+q2+q3

= 1.881 KJ +6.01 KJ + 4.514 KJ

= 12.405 KJ

5 0

3 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

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

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

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

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

3 0

3 years ago

The metal zirconium becomes superconducting at temperatures below 3.4000K.

dybincka [34]

Answer:

-269.75 degrees Celsius

Explanation:

7 0

3 years ago

When a radioactive sample decays for 2 half-lives the amount remaining will be ____ of the original?

mash [69]

1/4 of the original.

4 0

3 years ago

A 3.24-gram sample of NaHCO3 was completely decomposed in an experiment. 2NaHCO3 → Na2CO3 + H2CO3 In this experiment, carbon dio

Stolb23 [73]

Answer:

a) mass of the H2CO3 produced:

given:

Mass of sample = 3.24 g

Mass of Na2CO3 obtained after decomposition = 2.19 g

Solution :

Molar mass of NaHCO3 = 84

reaction:

2NaHCO3 → Na2CO3 + H2CO3

so it is clear that 2 mole of NaHCO3 gives 1 mole of Na2CO3 and H2CO3

Now, ICE table for the reaction is :

NaHCO3 Na2CO3 H2CO3

I 3.24/84 0 0

C -2x +x +x

E 3.24/84 -2x x x

As NaHCO3 is completely decomposed so final Concentration of NaHCO3 is zero.

=> 3.24/84 -2x = 0

=> 2x = 3.24/84

=> x = 1.62/84

The new ICE table is :

NaHCO3 Na2CO3 H2CO3

I 3.24/84 0 0

C -2x = -2(1.62/84) +x = 1.62/84 +x = 1.62/84

E 0 1.62/84 1.62/84

From the above ICE table,

it is found that (1.62/84 ) moles of H2CO3 is obtained.

Since,

The molar mass of H2CO3 is 62

=> Mass of H2CO3 obtained = moles × molar mass

=> Mass of H2CO3 obtained = (1.62 /84 ) × 62

= 1.19 grams

Mass of H2CO3 experimentally :

Mass of reactants = mass of products

=> Mass of sample = mass of Na2CO3 + mass of H2CO3

=> Mass of H2CO3 = mass of sample - mass of Na2CO3

= 3.24 - 2.19 = 1.05 g

b) Experimental mass = 1.05 g

Theoretical mass = 1.19 g

Percentage yield of H2CO3 = Experimental mass × 100 / Theoretical mass

= 1.05 × 100 /1.19

= 88.23 %

6 0

3 years ago