Answer is: mass of salt is 311,15 g.
V(H₂O) = 1,48 l · 1000 ml/l = 1480 ml.
m(H₂O) = 1480 g = 1,48 kg.
d(solution) = 1,00 g/ml.
ΔT(solution) = 13,4°C = 13,4 K.
Kf = 1,86 K·kg/mol; cryoscopic constant of water
i(NaCl) = 2; Van 't Hoff factor.
ΔT(solution) = Kf · b · i.
b(NaCl) = 13,4 K ÷ (1,86 K·kg/mol · 2).
b(NaCl) = 3,6 mol/kg.
n(NaCl) = 3,6 mol · 1,48 kg= 5,328 mol.
m(NaCl) = 5,328 mol · 58,4 g/mol = 311,15 g.
Given the concentration of aniline hydrochloride is 
Aniline hydrochloride is the conjugate acid of aniline a weak base.
pH can be calculated from 
anilinium ion the conjugate acid of aniline.
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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In acidic solutions you have H+ but in basic solutions you have OH-.
You need know that for to balance the reaction.
Volume of Argon V1 = 5.0 L
Pressure of Argon P1 = 2 atm
Final temperature T2 = 30 C = 30 + 273 = 303 K
Volume at final temperature V2= 6 L
Pressure at final temperature P2 = 8 atm
We know that (P1 x V1) / T1 = (P2 x V2) / T2
(2 x 5)/ T1 = (8 x 6)/ 303 => T1 = (10 x 303) / 48
Initial Temperature T1 = 3030 / 48 = 63.12
Initial Temperature = -209. 8 C
