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3 years ago

What is the empirical formula of an oxide of nitrogen containing 63.61% by mass of nitrogen and 36.69% by mass of oxygen?

Chemistry

1 answer:

katen-ka-za [31]3 years ago

8 0

Answer:

D. N₂O

Explanation:

Let's assume we have 100 g of the compound. That means it consists of 63.61 grams of nitrogen and 36.69 grams of oxygen.

Converting masses to moles:

63.61 g N × (1 mol N / 14.01 g N) = 4.540 mol N

36.69 g O × (1 mol O / 16.00 g O) = 2.293 mol O

Normalize by dividing by the smallest:

4.540 / 2.293 = 1.980 mol N

2.293 / 2.293 = 1.000 mol O

So there is approximately twice as many N atoms as O atoms. The empirical formula is therefore N₂O.

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Answer:

1,300,000,000,000

Explanation:

1.3 x 10^12

We want to convert this from scientific notation.

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2 years ago

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Solution :

The balanced cell reaction will be,

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Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half oxidation-reduction reaction will be :

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Now we have to calculate the Gibbs free energy.

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where,

$\Delta G^o$ = Gibbs free energy = ?

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F = Faraday constant = 96500 C/mole

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Now put all the given values in this formula, we get the Gibbs free energy.

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3 years ago

A mixture of 454 kg of applesauce at 10 degrees Celsius is heated in a heat exchanger by adding 121300 kJ. Calculate the outlet

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Explanation:

The given data is as follows.

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So, Heat given by heat exchanger = heat taken by apple sauce

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Thus, we can conclude that outlet temperature of the apple sauce is $76.5^{o}C$ .

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3 years ago

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3 years ago

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