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tatyana61 [14]
2 years ago
13

Ammonia gas combines with excess oxygen gas to produce nitric oxide and water. How many grams of ammonia gas would have to react

in order to release 154 kJ of energy
Chemistry
1 answer:
Sedbober [7]2 years ago
3 0

Answer:

11.6g of NH₃(g) have to react

Explanation:

For the reaction:

4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g) ΔH = -905kJ

<em>4 moles of ammonia produce 905kJ</em>

Thus, if you want to produce 154kJ of energy you need:

154kJ × (4 mol NH₃ / 905kJ) = <em>0.681moles of NH₃. </em>In mass -Molar mass ammonia is 17.031g/mol-

0.681mol NH₃ × (17.031g / mol) = <em>11.6g of NH₃(g) have to react</em>

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Which is the best way to represent 0.0035 kg by using scientific notation?
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A chemist prepares a solution of sodium thiosulfate by measuring out of sodium thiosulfate into a volumetric flask and filling t
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Mass of sodium thiosulfate Na_{2}S_{2}O_{3} is 110. g

Volume of the solution is 350. mL

Calculating the moles of sodium thiosulfate:

110. g Na_{2}S_{2}O_{3} * \frac{1 mol Na_{2}S_{2}O_{3}}{158.1 g Na_{2}S_{2}O_{3}} = 0.696 molNa_{2}S_{2}O_{3}

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3 years ago
Suppose an ice cube weighing 36.0 g at a temperature of 10°C is placed in 360 g water at a temperature of 20°C. Calculate the te
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Answer:

10.44 °C

Explanation:

When the thermal equilibrium is reached, both of the substances have the same final temperature (T). The liquid water will lose heat, and the ice cube will absorb this heat. The temperature of the ice will increase until it reaches 0°C, at this temperature, it will change of phase for liquid, absorbing heat, but without a change in the temperature. Then the temperature will increase until the equilibrium.

By the energy conservation, the total amount of heat must be equal to 0:

Qice + Qmelting + Qliquid1 + Qliquid2 = 0

Liquid 1 is the ice after melting, and liquid 2 the liquid that was already at the flask. When there's a change of temperature:

Q = n*c*ΔT, where n is the number of moles, c is the heat capacity and ΔT is the temperature change (final - initial). The temperature variation in °C is equal in K, so the temperature may be used in °C.

The melting heat is:

Q = n*Hfus, Hfus = 6007 J/mol

The molar mass of the water is 18 g/mol, so the number of moles of the water and the ice are:

nwater = nliquid1 = 360/18 = 20 moles

nice = 36/18 = 2 moles

Qice + Qmelting + Qliquid1 + Qliquid2 = 0

2*38*(0 - (-10)) + 2*6007 + 2*75*(T - 0) + 20*75*(T - 20) = 0

760 + 12014 + 150T + 1500T - 30000 = 0

1650T = 17226

T = 10.44 °C

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