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A queous solutions  حلول queous

In troduction:  في المقدمة:

base cistino  قاعدة السيستين
A solution is a homogenbus mixture ronsioting of a solute dissolved into a lsolvent. The solute is the substance. That is being dissolved, while the solvent istheldiosolving medium a) 0 g.e
المحلول عبارة عن خليط متجانس من مذاب مذاب في مذيب. المذاب هو المادة. يتم إذابة ذلك ، في حين أن المذيب isodeling media أ) 0 g.e
  1. A nonelectrolyte is a compound that doesen’t conduct an electric
    غير المنحل بالكهرباء هو مركب لا يوصل الكهرباء
IS current solution  الحل الحالي للنظام الدولي للبيانات
E x p ― ― = C 12 H 22 O 11 sp) ⟶ c 12 H 22 O 11 ( aq ) E x p _ _ = C 12 H 22 O 11  sp)  ⟶ c 12 H 22 O 11 ( aq ) Exp__=C_(12)H_(22)O_(11)^(" sp) ")longrightarrowc_(12)H_(22)O_(11(aq))\underline{\underline{E x p}}=C_{12} \mathrm{H}_{22} \mathrm{O}_{11} \stackrel{\text { sp) }}{ } \longrightarrow c_{12} \mathrm{H}_{22} \mathrm{O}_{11(\mathrm{aq})}
3 * An electrolyte is a compound that doesn’t conduct an electric current in agereus salution:
3 * المنحل بالكهرباء هو مركب لا يوصل تيارا كهربائيا في تحية agereus:
Exp: NaCl → H 2 O ( Na + , Cl − ) (aa) KCl ( s ) ⟶ ( K + + φ − ) ( a 4 )  Exp:  NaCl → H 2 O Na + , Cl − (aa)  KCl ( s ) ⟶ K + + φ − a 4 {:[" Exp: "NaClrarr"H_(2)O"(Na^(+),Cl^(-))_((aa) )],[KCl(s)longrightarrow(K^(+)+varphi^(-))(a_(4))]:}\begin{aligned} \text { Exp: } & \mathrm{NaCl} \xrightarrow{\mathrm{H}_{2} \mathrm{O}}\left(\mathrm{Na}^{+}, \mathrm{Cl}^{-}\right)_{\text {(aa) }} \\ & \mathrm{KCl(s)} \longrightarrow\left(\mathrm{~K}^{+}+\varphi^{-}\right)\left(\mathrm{a}_{4}\right) \end{aligned}
  • A strong electiolyte is asolution in which almost all of the dísolved solations solute exists as tro ions NaCl → Na + + Cl − NaCl → Na + + Cl − NaClrarrNa^(+)+Cl^(-)\mathrm{NaCl} \rightarrow \mathrm{Na}^{+}+\mathrm{Cl}^{-}
    الكهربية القوية هي الحل الذي توجد فيه جميع الحلول المذابة تقريبا على شكل أيونات NaCl → Na + + Cl − NaCl → Na + + Cl − NaClrarrNa^(+)+Cl^(-)\mathrm{NaCl} \rightarrow \mathrm{Na}^{+}+\mathrm{Cl}^{-}
  • A weak electrolyte is a solotion in which only a small fraction of the dissolved solute exists as ions HNO 2 ⇆ H + + NO 2 2 − HNO 2 ⇆ H + + NO 2 2 − HNO_(2)⇆H^(+)+NO_(2)^(2-)\mathrm{HNO}_{2} \leftrightarrows \mathrm{H}^{+}+\mathrm{NO}_{2}^{2-}
    المنحل بالكهرباء الضعيف هو منفرد لا يوجد فيه سوى جزء صغير من المذاب المذاب على شكل أيونات HNO 2 ⇆ H + + NO 2 2 − HNO 2 ⇆ H + + NO 2 2 − HNO_(2)⇆H^(+)+NO_(2)^(2-)\mathrm{HNO}_{2} \leftrightarrows \mathrm{H}^{+}+\mathrm{NO}_{2}^{2-}
Poncentration of solutions.
تركيز الحلول.
Molar concontration (Molarity) : is defined asthe mumbre of voles of cssition solute devided by the volume of colution:
concontration Molar (Molarity) : يتم تعريفه على أنه Mumbre من فئران cssition ذائب مقسمة بحجم colution:
c = n ( solute ) V ( solution ) : ( mal / L ) ( M ) c = n (  solute  ) V (  solution  ) : (  mal  / L ) ( M ) c=(n(" solute "))/(V(" solution ")):(" mal "//L)(M)c=\frac{n(\text { solute })}{V(\text { solution })}:(\text { mal } / L)(M)
weight concentration , is defined as the mass of the colofe devided by the volume of solution:
يتم تعريف تركيز الوزن على أنه كتلة الكولوف المقسمة بحجم المحلول:
c w = m ( solvte ) V (solmtion) lg ⁡ / φ ) c w = m ⋅ M V = C ⋅ M / n = m M ⇒ m = n ⋅ M c w = m (  solvte  ) V  (solmtion)  lg ⁡ / φ c w = m ⋅ M V = C ⋅ M / n = m M ⇒ m = n ⋅ M {:[{:c_(w)=(m(" solvte "))/(V" (solmtion) ")lg //varphi)],[c_(w)=(m*M)/(V)=C*M quad//n=(m)/(M)=>m=n*M]:}\begin{aligned} & \left.c_{w}=\frac{m(\text { solvte })}{V \text { (solmtion) }} \lg / \varphi\right) \\ & c_{w}=\frac{m \cdot M}{V}=C \cdot M \quad / n=\frac{m}{M} \Rightarrow m=n \cdot M \end{aligned}
Remembre,  تذكر،
1 dm = 1 10 × 100 cm = 10 cm 1 l = 10 1 cm 3 / m 3 = 10 3 l 1 dm = 1 10 × 100 cm = 10 cm 1 l = 10 1 cm 3 / m 3 = 10 3 l {:[1dm=(1)/(10)xx100cm=10cm],[1l=10^(1cm^(3))//m^(3)=10^(3)l]:}\begin{aligned} & 1 \mathrm{dm}=\frac{1}{10} \times 100 \mathrm{~cm}=10 \mathrm{~cm} \\ & 1 \mathrm{l}=10^{1 \mathrm{~cm}^{3}} / \mathrm{m}^{3}=10^{3} \mathrm{l} \end{aligned}
molar fraction: ρ := n i ∑ n i ρ := n i ∑ n i rho:=(n_(i))/(sumn_(i))\rho:=\frac{n_{i}}{\sum n_{i}}
الكسر المولي: ρ := n i ∑ n i ρ := n i ∑ n i rho:=(n_(i))/(sumn_(i))\rho:=\frac{n_{i}}{\sum n_{i}}
Specific gravity: S.G Guguldä̃l
الثقل النوعي: S.G Guguldä̃l
S G = ρ sobstance ρ nefrence (does not have imits) S G = ρ sub ρ H 2 O ρ → density aitwit ρ H 2 O → 1 g / 1000 kg / cm 2 ρ = m v kg / m 3 2lat aiozol S G = ρ sobstance  ρ nefrence   (does not have imits)  S G = ρ  sub  ρ H 2 O ρ →  density aitwit  ρ H 2 O → 1 g / 1000 kg / cm 2 ρ = m v kg / m 3  2lat   aiozol  {:[SG=(rho_("sobstance "))/(rho_("nefrence "))" (does not have imits) "],[SG=(rho" sub ")/(rho_(H_(2)O))],[rho rarr" density aitwit "],[rho_(H_(2)O)rarr1g//1000kg//cm^(2)],[rho=(m)/(v)kg//m^(3)],[" 2lat "],[" aiozol "]:}\begin{aligned} & S G=\frac{\rho_{\text {sobstance }}}{\rho_{\text {nefrence }}} \text { (does not have imits) } \\ & S G=\frac{\rho \text { sub }}{\rho_{\mathrm{H}_{2} \mathrm{O}}} \\ & \rho \rightarrow \text { density aitwit } \\ & \rho_{H_{2} \mathrm{O}} \rightarrow 1 \mathrm{~g} / 1000 \mathrm{~kg} / \mathrm{cm}^{2} \\ & \rho=\frac{m}{v} \mathrm{~kg} / \mathrm{m}^{3} \\ & \text { 2lat } \\ & \text { aiozol } \end{aligned}
If S G > 1 S G > 1 SG > 1S G>1 the object will sink in th neference substance if S G < 1 S G < 1 SG < 1S G<1 the object will floatin the sip:-
إذا كان S G > 1 S G > 1 SG > 1S G>1 الجسم سوف يغرق في المادة المرجعية إذا S G < 1 S G < 1 SG < 1S G<1 كان الجسم سوف يطفو في الرشفة: -
mole fraction, Nodramsin
كسر الخلد ، نودرامسين
f 1 f 1 f_(1)f_{1} : Mole Fraction
كسر الخلد f 1 f 1 f_(1)f_{1} :
n 1 n 1 n_(1)n_{1} : so lute moles
n 1 n 1 n_(1)n_{1} حتى العود الشامات
f 2 f 2 f_(2)f_{2} : Mole Fraction
كسر الخلد f 2 f 2 f_(2)f_{2} :
n 2 n 2 n_(2)n_{2} : solvent moles
مولات n 2 n 2 n_(2)n_{2} المذيبات:
f 1 = n 1 n 1 + n 2 f 1 = n 1 n 1 + n 2 f_(1)=(n_(1))/(n_(1)+n_(2))f_{1}=\frac{n_{1}}{n_{1}+n_{2}}
f 1 + f 2 = 1 f 1 + f 2 = 1 f_(1)+f_(2)=1f_{1}+f_{2}=1
f 2 = n 2 n 1 + n 2 f 2 = n 2 n 1 + n 2 f_(2)=(n_(2))/(n_(1)+n_(2))f_{2}=\frac{n_{2}}{n_{1}+n_{2}}
  • Parts pear million  أجزاء الكمثرى مليون
    p.p.m = m solut liteorofsolution × 10 6 = m  solut   liteorofsolution  × 10 6 =(m" solut ")/(" liteorofsolution ")xx10^(6)=\frac{m \text { solut }}{\text { liteorofsolution }} \times 10^{6}  بعد = m solut liteorofsolution × 10 6 = m  solut   liteorofsolution  × 10 6 =(m" solut ")/(" liteorofsolution ")xx10^(6)=\frac{m \text { solut }}{\text { liteorofsolution }} \times 10^{6} الظهر
  • P.p.b (part’s per billion)
    P.p.b (جزء لكل مليار)
= m litire ofsolution × 10 9 = m  litire ofsolution  × 10 9 =(m)/(" litire ofsolution ")xx10^(9)=\frac{m}{\text { litire ofsolution }} \times 10^{9}
Osmolanity ( 0 )  الأسمولانيتي ( 0 )
The osmolarity of solution is a measune of the soncetration of solute in the solution with (osmol /L)
الأسمولية من المحلول هو قياس من ال [سإكستر] من المذاب في المحلول مع ([أوسمول] / [ل])
we have i = w c i = w c i=(w)/(c)i=\frac{w}{c}  لدينا i = w c i = w c i=(w)/(c)i=\frac{w}{c}
a) for the solutions non électrolyte (wrec and glucose C 4 H 4 O 2 C 4 H 4 O 2 C_(4)H_(4)O_(2)C_{4} H_{4} \mathrm{O}_{2} ) α = 0 ⇒ i = w c = 1 ⇒ w = c α = 0 ⇒ i = w c = 1 ⇒ w = c alpha=0=>i=(w)/(c)=1=>w=c\alpha=0 \Rightarrow i=\frac{w}{c}=1 \Rightarrow w=c
أ) للمحاليل non électrolyte (wrec and glucose C 4 H 4 O 2 C 4 H 4 O 2 C_(4)H_(4)O_(2)C_{4} H_{4} \mathrm{O}_{2} ) α = 0 ⇒ i = w c = 1 ⇒ w = c α = 0 ⇒ i = w c = 1 ⇒ w = c alpha=0=>i=(w)/(c)=1=>w=c\alpha=0 \Rightarrow i=\frac{w}{c}=1 \Rightarrow w=c
b) for the electrolyte weak ( 0 < α < 1 ) ( 0 < α < 1 ) (0 < alpha < 1)(0<\alpha<1)
ب) للإلكتروليت ضعيف ( 0 < α < 1 ) ( 0 < α < 1 ) (0 < alpha < 1)(0<\alpha<1)

Exp?  اكسب؟

(1) t = 0 A B → A + + B − t c 0 t ( 1 − α ) c α c α c t = 0 A B → A + + B − t c 0 t ( 1 − α ) c α c α c {:[t=0,AB rarrA^(+)+B^(-)],[t quad,c,0],[t,(1-alpha)c,alpha c quad alpha c]:}\begin{array}{ccc}t=0 & A B \rightarrow A^{+}+B^{-} \\ t \quad & c & 0 \\ t & (1-\alpha) c & \alpha c \quad \alpha c\end{array}
so w = ( 1 − α ) c + α c + α c = c − α c + α c + α c = c + α c = ( 1 + α ) c and i = 1 + α  so  w = ( 1 − α ) c + α c + α c = c − α c + α c + α c = c + α c = ( 1 + α ) c  and  i = 1 + α " so "{:[w=(1-alpha)c+alpha c+alpha c],[=c-alpha c+alpha c+alpha c],[=c+alpha c=(1+alpha)c" and "i=1+alpha]:}\text { so } \begin{aligned} w & =(1-\alpha) c+\alpha c+\alpha c \\ & =c-\alpha c+\alpha c+\alpha c \\ & =c+\alpha c=(1+\alpha) c \text { and } i=1+\alpha \end{aligned}
(2). A B 2 = A + + + 2 B − A B 2 = A + + + 2 B − AB_(2)=A^(++)+2B^(-)A B_{2}=A^{++}+2 B^{-}
t = c 0 0 ( 1 − α ) c α c 2 α c w = ( 1 − α ) c + α c + 2 α c = c − 2 c + α c + 2 α c = c + 2 α c = ( 1 + 2 α ) c and i = 1 + 2 α t = c 0 0 ( 1 − α ) c α c 2 α c w = ( 1 − α ) c + α c + 2 α c = c − 2 c + α c + 2 α c = c + 2 α c = ( 1 + 2 α ) c  and  i = 1 + 2 α {:[t=c quad0quad0],[(1-alpha)c quad alpha c quad2alpha c],[w=(1-alpha)c+alpha c+2alpha c],[=c-2c+alpha c+2alpha c],[=c+2alpha c=(1+2alpha)c" and "i=1+2alpha]:}\begin{aligned} t= & c \quad 0 \quad 0 \\ & (1-\alpha) c \quad \alpha c \quad 2 \alpha c \\ w= & (1-\alpha) c+\alpha c+2 \alpha c \\ = & c-2 c+\alpha c+2 \alpha c \\ = & c+2 \alpha c=(1+2 \alpha) c \text { and } i=1+2 \alpha \end{aligned}