S4O5
S3O
SeF6
N4S5
CCl9
All numbers should be subscripts
Answer:
12 moles of CO₂.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
CO₂ + H₂O —> H₂CO₃
From the balanced equation above,
1 mole of CO₂ dissolves in water to produce 1 mole of H₂CO₃.
Finally, we shall determine the number of moles of CO₂ that will dissolve in water to produce 12 moles of H₂CO₃. This can be obtained as follow:
From the balanced equation above,
1 mole of CO₂ dissolves in water to produce 1 mole of H₂CO₃.
Therefore, 12 moles of CO₂ will also dissolve in water to produce 12 moles of H₂CO₃.
Thus, 12 moles of CO₂ is required.
From the mole ratio of the reaction as given in the equation of the reaction, the mass of calcium chloride that can be produced from 5.59 mol of hydrochloric acid is 310.245 g.
<h3>What mass of calcium chloride can be produced from 5.59 mol of hydrochloric acid?</h3>
The mass of calcium chloride that can be produced from 5.59 mol of hydrochloric acid is determined from the equation of the reaction.
The equation of the reaction is given below:
Ca²⁺ (aq) + 2 HCl (aq) ---> CaCl₂ (s) + H₂ (g)
From the equation of the reaction, the mole ratio of HCL and calcium chloride is 2 : 1
Therefore, moles of calcium chloride that can be produced will be:
The moles of calcium chloride = 5.59 moles * 1/2
The moles of calcium chloride = 2.795 moles
The mass of calcium chloride produced = moles * molar mass
Molar mass of calcium chloride = 111 g/mol
Mass of calcium chloride produced = 2.795 * 111
Mass of calcium chloride produced = 310.245 g
Learn more about mole ratio at: brainly.com/question/19099163
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I believe it's about the dislodging response as Ag structures solvent salt with nitrate. For dislodging response, more receptive metal will uproot less receptive metal from the arrangement. In your question, the reactivity of the metal are positioned as takes after: Mg > Cu > Ag. Take note of that more responsive metals, that are Mg and Cu, are in the arrangement.
Answer:
hope it helps you...
Explanation:
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