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3 years ago

List two possible steps where impurities may be gained, which would appear to be an increase in copper

Chemistry

1 answer:

Arisa [49]3 years ago

8 0

Answer:

i) During washing

ii) During Measurement

Explanation:

The two possible steps are :

i) During washing : during the washing method the residue may be not completely dried out and this residue ( water) will add up to the final product ( copper yield ) and this kind of error is called human error.

ii) During measurement : If the weighing instrument is faulty there might be addition in value of the final copper yield which will see the final yield value > 100% . this error occurs when the initial value and final value is been weighed

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Calculate the density of O2(g) at 415 K and 310 bar using the ideal gas and the van der Waals equations of state. Use a numerica

Lera25 [3.4K]

Answer:

Explanation:

From the given information:

The density of O₂ gas = $d_{ideal} = \dfrac{P\times M}{RT}$

here:

P = pressure of the O₂ gas = 310 bar

= $310 \ bar \times \dfrac{0.987 \ atm}{1 \ bar}$

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The rate R = 0.0821 L.atm/mol.K

molar mass of O₂ gas = 32 g/mol

∴

$d_{ideal} = \dfrac{305.97 \ \times 32}{0.0821 \times 415}$

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To find the density using the Van der Waal equation

Recall that:

the Van der Waal constant for O₂ is:

a = 1.382 bar. L²/mol² &

b = 0.0319 L/mol

The initial step is to determine the volume = Vm

The Van der Waal equation can be represented as:

$P =\dfrac{RT}{V-b}-\dfrac{a}{V^2}$

where;

R = gas constant (in bar) = 8.314 × 10⁻² L.bar/ K.mol

Replacing our values into the above equation, we have:

$310 =\dfrac{0.08314\times 415}{V-0.0319}-\dfrac{1.382}{V^2}$

$310 =\dfrac{34.5031}{V-0.0319}-\dfrac{1.382}{V^2}$

$310V^3 -44.389V^2+1.382V-0.044=0$

After solving;

V = 0.1152 L

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$d_{Van \ der \ Waal}$ = 277.77 g/L

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lbvjy [14]

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vivado [14]

Answer:

Explanation:

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