Answer:
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The mass in grams of NH₃ produced from the reaction is 3.4 g
<h3>Balanced equation</h3>
We'll begin by writing the balanced equation for the reaction. This illustrated below:
N₂ + 3H₂ -> 2NH₃
From the balanced equation above,
1 dm³ of N₂ reacted to produced 2 dm³ NH₃
<h3>How to determine the volume of NH₃ produced</h3>
From the balanced equation above,
1 dm³ of N₂ reacted to produced 2 dm³ NH₃
Therefore,
2.24 dm³ of N₂ will react to produce = 2.24 × 2 = 4.48 dm³ of NH₃
<h3>How to determine the mass of NH₃ produced</h3>
We'll begin by obtained the mole of 4.48 dm³ of NH₃. Details below:
22.4 dm³ = 1 mole NH₃
Therefore,
4.48 dm³ = 4.48 / 22.4
4.48 dm³ = 0.2 mole of NH₃
Finally, we shall determine the mass of NH₃ as follow:
- Molar mass of NH₃ = 17 g/mol
- Mole of NH₃ = 0.2 mole
- Mass of NH₃ =?
Mass = mole × molar mass
Mass of NH₃ = 0.2 × 17
Mass of NH₃ = 3.4 g
Learn more about stoichiometry:
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Answer:
endothermic reactions
Explanation:
Endothermic reactions are those reactions that absorbs heat from the surroundings as opposed to the exothermic reaction that releases heat energy. This heat absorption that characterizes endothermic reactions causes the surrounding to be cool.
However, in an endothermic reaction, more energy is needed to be absorbed in order to break the bonds of the reactants to form products, hence, making the activation energy of endothermic reactions HIGHER.
Answer:
A. boiling point
Explanation:
The correct answer is;
A. boiling point
The boiling point of a liquid is the temperature at which the vapour pressure is equal to the atmospheric pressure.
B. melting point
This option is wrong because the melting point is the temperature at which a solid undergoes a phase change to a liquid.
C. freezing point
This option is wrong because the freezing point is the temperature at which a liquid undergoes a phase change to a solid.
Answer: There are 0.0637 moles present in 85.0 mL of 0.750 M KOH.
Explanation:
Given: Volume = 85.0 mL (1 mL = 0.001 L) = 0.085 L
Molarity = 0.750 M
It is known that molarity is the number of moles of solute present in liter of a solution.
Therefore, moles present in given solution are calculated as follows.
Thus, we can conclude that there are 0.0637 moles present in 85.0 mL of 0.750 M KOH.
