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Leviafan [203]
2 years ago
5

Formula of a copper (II)sulfate hydrate lab

Chemistry
1 answer:
s344n2d4d5 [400]2 years ago
7 0

Answer:

Weigh the empty crucible, and then weigh into it between 2 g and 3 g of hydrated copper(II) sulphate. Record all weighings accurate to the nearest 0.01 g.

Support the crucible securely in the pipe-clay triangle on the tripod over the Bunsen burner.

Heat the crucible and contents, gently at first, over a medium Bunsen flame, so that the water of crystallisation is driven off steadily. The blue colour of the hydrated compound should gradually fade to the greyish-white of anhydrous copper(II) sulfate. Avoid over-heating, which may cause further decomposition, and stop heating immediately if the colour starts to blacken. If over-heated, toxic or corrosive fumes may be evolved. A total heating time of about 10 minutes should be enough.

Allow the crucible and contents to cool. The tongs may be used to move the hot crucible from the hot pipe-clay triangle onto the heat resistant mat where it should cool more rapidly.

Re-weigh the crucible and contents once cold.

Calculation:

Calculate the molar masses of H2O and CuSO4 (Relative atomic masses: H=1, O=16, S=32, Cu=64)

Calculate the mass of water driven off, and the mass of anhydrous copper(II) sulfate formed in your experiment

Calculate the number of moles of anhydrous copper(II) sulfate formed

Calculate the number of moles of water driven off

Calculate how many moles of water would have been driven off if 1 mole of anhydrous copper(II) sulfate had been formed

Write down the formula for hydrated copper(II) sulfate.

#*#*SHOW FULLSCREEN*#*#

Explanation:

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Which statement refers to the quantum mechanical model? A) Electrons move in a circular path about the nucleus. B) Electrons can
Yuri [45]

Answer:

Electrons are in "orbitals", regions of space where there is high probability of being found.

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3 years ago
I need help on both a and b of question 1
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Answer:

(a) -0.00017 M/s;

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Explanation:

(a) Rate of a reaction is defined as change in molarity in a unit time, that is:

r = \frac{\Delta c}{\Delta t}

Given the following reaction:

2 N_2O_5 (g)\rightleftharpoons 4 NO_2 (g) + O_2 (g)

We may write the rate expression in terms of reactants firstly. Since reactants are decreasing in molarity, we're adding a negative sign. Similarly, if we wish to look at the overall reaction rate, we need to divide by stoichiometric coefficients:

r = -\frac{\Delta [N_2O_5]}{2 \Delta t}

Reaction rate is also equal to the rate of formation of products divided by their coefficients:

r = \frac{\Delta [NO_2]}{4 \Delta t} = \frac{\Delta [O_2]}{\Delta t}

Let's find the rate of disappearance of the reactant firstly. This would be found dividing the change in molarity by the change in time:

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(b) Using the relationship derived previously, we know that:

-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}

Rate of appearance of nitrogen dioxide is given by:

r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t}

Which is obtained from the equation:

-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}

If we multiply both sides by 4, that is:

-\frac{4 \Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{\Delta t}

This yields:

[tex]r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t} = -2\frac{\Delta [N_2O_5]}{ \Delta t} = -2\cdot (-0.00017 M/s) = 0.00034 M/s[tex]

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Calculating for moles O given mass:

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Answer:

<span>0.2 moles O</span>

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