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mariarad [96]
3 years ago
9

How many grams of NH3 can be produced from 2.30 mol of N2 and excess H2.

Chemistry
1 answer:
MrRa [10]3 years ago
3 0
<h3>Answer:</h3>

78.34 g

<h3>Explanation:</h3>

From the question we are given;

Moles of Nitrogen gas as 2.3 moles

we are required to calculate the mass of NH₃ that may be reproduced.

<h3>Step 1: Writing the balanced equation for the reaction </h3>

The Balanced equation for the reaction is;

   N₂(g) + 3H₂(g) → 2NH₃(g)

<h3>Step 2: Calculating the number of moles of NH₃</h3>

From the equation 1 mole of nitrogen gas reacts to produce 2 moles of NH₃

Therefore, the mole ratio of N₂ to NH₃ is 1 : 2

Thus, Moles of NH₃ = Moles of N₂ × 2

                                  = 2.3 moles × 2

                                  = 4.6 moles

<h3>Step 3: Calculating the mass of ammonia produced </h3>

Mass = Moles × molar mass

Molar mass of ammonia gas = 17.031 g/mol

Therefore;

Mass = 4.6 moles × 17.031 g/mol

         = 78.3426 g

         = 78.34 g

Thus, the mass of NH₃ produced is 78.34 g

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Answer:My answer is in the photo

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8 0
3 years ago
A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The
True [87]
First, in order to calculate the specific heat capacity of the metal in help in identifying it, we must find the heat absorbed by the calorimeter using:
Energy = mass * specific heat capacity * change in temperature
Q = 250 * 1.035 * (11.08 - 10)
Q = 279.45 cal/g

Next, we use the same formula for the metal as the heat absorbed by the calorimeter is equal to the heal released by the metal.

-279.45 = 50 * c * (11.08 - 45) [minus sign added as energy released]
c = 0.165

The specific heat capacity of the metal is 0.165 cal/gC
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How many molecules are in 165 g of carbon dioxide?
Alexeev081 [22]
First you need to know how many moles you would get from 165 g of carbon dioxide.
Carbon = 12
Oxygen 2 = 16 x 2 = 32
Total: 32 + 12 = 44
- Use proportions
1 mol CO2 x 1 mol CO2 44 g
---------------- = ---------- or ----------------- = ----------
44 g 165 x 165 g

165/44 = 3.75 moles CO2
- Last step is to multiply the amount of moles by Avagadros number, 6.02 x 10^23
3.75 x 6.02 x 10^23 = 2.25 x 10^24 molecules

- Keep in mind if your answer can be a little different than mind if you either round or keep the same numbers. I hope this Really Helps ! c:
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Br2(l) + 2Nal(aq) — 12(s) + 2NaBr(aq)
Ganezh [65]

Iodine (I) is oxidized, and bromine (Br) is reduced. The correct option is the last option - lodine (I) is oxidized, and bromine (Br) is reduced.

To determine which elements are oxidized and reduced,

First, we will define the terms Oxidation and Reduction

<em>Oxidation </em>is simply defined as the loss of electrons. It can also be defined as increase in oxidation number.

<em>Reduction </em>is defined as the gain of electrons. It can also be defined as decrease in oxidation number.

The given chemical equation is

Br₂(l) + 2NaI(aq) → I₂(s) + 2NaBr(aq)

Oxidation number of Bromine decreased from 0 to -1.

Therefore, Bromine is reduced.

Oxidation number of Iodine increased from -1 to 0.

Therefore, Iodine is oxidized.

Oxidation number of sodium did not change.

Therefore, Sodium is neither oxidized nor reduced.

Hence, Iodine (I) is oxidized, and bromine (Br) is reduced. The correct option is the last option - lodine (I) is oxidized, and bromine (Br) is reduced.

Learn more here: brainly.com/question/12913997

3 0
2 years ago
2. You have a 2.50 mole gas sample in a 500.0 mL flask at 25.0 °C.
SIZIF [17.4K]

Considering the ideal gas law, the pressure of the gas sample is 122.18 atm.

<h3>What is an ideal gas</h3>

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

<h3>Definition of ideal gas law</h3>

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of gases:

P×V = n×R×T

<h3>Pressure of the gas sample</h3>

In this case, you know:

  • P= ?
  • V= 500 mL= 0.5 L
  • n= 2.50 moles
  • R= 0.082 \frac{atm L}{mol K}
  • T= 25 °C= 298 K

Replacing in the ideal gas law:

P×0.5 L = 2.50 moles ×0.082 \frac{atm L}{mol K} ×298 K

Solving:

P= (2.50 moles ×0.082 \frac{atm L}{mol K} ×298 K)÷ 0.5 L

<u><em>P= 122.18 atm</em></u>

Finally, the pressure of the gas sample is 122.18 atm.

Learn more about the ideal gas law:

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