The equation of the reaction before balancing is
a0NH₄Cl + a1Ag₃PO₄ → a2AgCl + a3(NH₄)₃PO₄
PO₄³⁻ ion is balanced.
on the left side, theres 1 (NH₄⁺) ion and right side 3 (NH₄⁺) ions. Therefore if we put the coefficient for NH₄Cl, we will obtain the following equation
3 NH₄Cl + a1Ag₃PO₄ → a2AgCl + a3(NH₄)₃PO₄
3 Ag⁺ ions on the left side and 1 Ag⁺ ion on the right side, so if we put the coefficient of AgCl as 3, following equation obtained
3 NH₄Cl + a1Ag₃PO₄ → 3 AgCl + a3(NH₄)₃PO₄
Cl⁻ ions are also balanced now, 3 on either side.
a1 and a3 are 1 as those compounds are as it is, so coefficient is 1 for both
balanced equation is as follows
3 NH₄Cl + Ag₃PO₄ → 3 AgCl + (NH₄)₃PO₄
coefficients are
a0 - 3
a1 - 1
a2 - 3
a3 - 1
Answer:
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1374.75 is the concentration in milligrams per ml of a solution containing 23.5 meq sodium chloride per milliliter.
Concentration in chemistry is calculated by dividing a constituent's abundance by the mixture's total volume.
It is calculated in mg/ml.
The unit of measurement frequently used for electrolytes is the milliequivalent (mEq). This value compares an element's chemical activity, or combining power, to that of 1 mg of hydrogen.
Formula for calculating concentration in mg/ml is
Conc. (mg/ml) = M(eq) /ml × Molecular weight / Valency
Given
M(eq) NaCl/ ml = 23.5
Molecular weight pf NaCl = 58.5 g/mol
Valency = 1
Putting the values into the formula
Conc. (mg/ml) = 23.5 ×58.5/1
= 1374.75 mg/ml
Hence, 1374.75 is the concentration in milligrams per ml of a solution containing 23.5 meq sodium chloride per milliliter.
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The % yield of Ca(OH)₂ : 62.98%
<h3>Further eplanation
</h3>
Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated
General formula:
Percent yield = (Actual yield / theoretical yield )x 100%
An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients
Reaction
CaO + H₂O ⇒ Ca(OH)₂
mass CaO= 4.2 g
mol CaO(MW=56,0774 g/mol) :
mol Ca(OH)₂ based on mol CaO
mol ratio CaO : Ca(OH)₂,= 1 : 1, so mol Ca(OH)₂ = 0.075
mass Ca(OH)₂(MW=74,093 g/mol) ⇒ theoretical
% yield :
