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alexgriva [62]
2 years ago
6

3. A 0.500 g sample of nitrogen gas combines with 1.140 g of oxygen gas to form NO2. If the atomic mass of oxygen is 16.000, cal

culate the atomic mass of nitrogen from this data.
Chemistry
2 answers:
DedPeter [7]2 years ago
6 0

Answer;

= 18.24

Explanation;

The ratio of N and O in the formula NO2 IS 1:2

Mass of nitrogen gas is 0.500 g

Moles of nitrogen will be;

= 0.500/16 = 0.03125 moles

Therefore;

The moles of Oxygen from the ratio will be;

= 0.03125 × 2 = 0.0625 moles

But; 0.0625 moles is equal to 1.140 g of Oxygen

The atomic number (mass in 1 mole) will be;

= 1.140 /0.0625

= 18.24

Thus the atomic number of Oxygen from the data is 18.24

astra-53 [7]2 years ago
6 0

The balanced chemical equation for the given reaction is:

N₂ + 2O₂ → 2NO₂

According to the given balanced equation, 2 moles of O₂ combines with 1 mole of O₂

Given, Mass of O₂ = 1.140 g

Atomic mass of O = 16 amu

Molar mass of O₂ = 16 x 2 = 32 g/mol

Now to calculate the moles of N₂:

1.14 g of O₂ x (1 mole of O₂/ 32 g of O₂) x (1 mol of N₂/ 2 mol O₂) = 0.0178 mol of N₂

Molar mass = mass/ moles  

Given, Mass of N₂ = 0.500 g

Molar mass of N₂ = 0.500 g / 0.0178 mol = 28 g/mol

Atomic mass of N = 28/2 = 14 amu

Therefore, the atomic mass of N is 14 amu


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When 1.04g of cyclopropane was burnt in excess oxygen in a bomb calorimeter, the temperature rose by 3.69K. The total heat capac
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Answer:

\Delta _{comb}H=-2,093\frac{kJ}{mol}

Explanation:

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In this case, since these calorimetry problems are characterized by the fact that the calorimeter absorbs the heat released by the combustion of the substance, we can write:

Q_{rxn}+Q_{cal}=0

Thus, given the temperature change and the total heat capacity, we obtain the following total heat of reaction:

Q_{rxn}=-14.01kJ/K*3.69K\\\\Q_{rxn}=-51.70kJ

Now, by dividing by the moles in 1.04 g of cyclopropane (42.09 g/mol) we obtain the enthalpy of combustion of this fuel:

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4 0
1 year ago
Use the following half-reactions to design a voltaic cell: Sn4+(aq) + 2 e− → Sn2+(aq) Eo = 0.15 V Ag+(aq) + e−→ Ag(s) Eo = 0.80
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E° = 0.65 V

Explanation:

Let's consider the following reductions and their respective standard reduction potentials.

Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V

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The reaction with the highest reduction potential will occur as a reduction while the other will occur as an oxidation. The corresponding half-reactions are:

Reduction (cathode): Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V

Oxidation (anode): Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻ E°red = 0.15 V

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Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2N
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Answer : The correct option is, (a) 0.44

Explanation :

First we have to calculate the concentration of N_2O_4.

\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

Now we have to calculate the dissociated concentration of N_2O_4.

The balanced equilibrium reaction is,

                             N_2O_4(g)\rightleftharpoons 2NO_2(aq)

Initial conc.           1.0 M          0

At eqm. conc.     (1.0-x) M    (2x) M

As we are given,

The percent of dissociation of N_2O_4 = \alpha = 28.0 %

So, the dissociate concentration of N_2O_4 = C\alpha=1.0M\times \frac{28.0}{100}=0.28M

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Now we have to calculate the concentration of N_2O_4\text{ and }NO_2 at equilibrium.

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Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

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

K_c=\frac{(0.56)^2}{0.72}=0.44

Therefore, the equilibrium constant K_c for the reaction is, 0.44

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