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Ad libitum [116K]
2 years ago
15

A solution containing a mixture of 0.0333 M0.0333 M potassium chromate ( K2CrO4K2CrO4 ) and 0.0532 M0.0532 M sodium oxalate ( Na

2C2O4Na2C2O4 ) was titrated with a solution of barium chloride ( BaCl2BaCl2 ) for the purpose of separating CrO2−4CrO42− and C2O2−4C2O42− by precipitation with the Ba2+Ba2+ cation. The solubility product constants ( KspKsp ) for BaCrO4BaCrO4 and BaC2O4BaC2O4 are 2.10×10−102.10×10−10 and 1.30×10−61.30×10−6 , respectively.
A) Which barium salt will precipitate first? O BaC204 O BaCrO4
B) What concentration of Ba2+ must be present for BaCrO4 to begin precipitating? Number [Ba2+]=
C) What concentration of Ba is required to reduce oxalate to 10% of its original concentration? Number 2+
D) What is the ratio of oxalate to chromate (IC204CrO4) when the Ba2 concentration is 0.0050 M? Number C,O Cro,
Chemistry
1 answer:
jasenka [17]2 years ago
3 0

Answer:

Explanation:

K₂CrO₄ + ( COONa )₂ + 2BaCl₂ = Ba CrO₄ + ( COO ) ₂ Ba + 2 KCl + 2 NaCl

.033 M             .053 M    

Ksp of   Ba CrO₄  is      2.10×10⁻¹⁰

Ksp of ( COO ) ₂ Ba  is   1.30×10⁻⁶

A ) Ksp of   Ba CrO₄ is less so it will precipitate out first .

B) Ksp = 2.10×10⁻¹⁰

Ba CrO₄ = Ba⁺²  +  CrO₄⁻²

                   C            .033

C x   .033  = 2.10×10⁻¹⁰

C =  63.63 x 10⁻¹⁰ M

Ba⁺² must be present in concentration = 63.63 x 10⁻¹⁰ M

C)

90% of precipitation of barium oxalate

concentration of oxalate to precipitate out = .9 x .0532 = .04788

( COO ) ₂ Ba  =  (COO)₂⁻²     +       Ba⁺²

                           .04788 M            C  

C x  .04788  =  1.30×10⁻⁶

C = 27.15  x 10⁻⁶ M .

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3 years ago
Potassium -39 has how many neutrons
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Potassium has 20 neutrons

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5 0
2 years ago
A 0.450 g sample of solid lead(II) nitrate is added to 250 mL of 0.250 M sodium iodide solution. Assume no change in volume of t
Verdich [7]

Pb(NO₃)₂ ⇒limiting reactant

moles PbI₂ = 1.36 x 10⁻³

% yield  = 87.72%

<h3>Further explanation</h3>

Given

Reaction(unbalanced)

Pb(NO₃)₂(s) + NaI(aq) → PbI₂(s) + NaNO₃(aq)

Required

  • moles of PbI₂
  • Limiting reactant
  • % yield

Solution

Balanced equation :

Pb(NO₃)₂(s) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)

mol Pb(NO₃)₂ :

= 0.45 : 331 g/mol

= 1.36 x 10⁻³

mol NaI :

= 250 ml x 0.25 M

= 0.0625

Limiting reactant (mol : coefficient)

Pb(NO₃)₂ : 1.36 x 10⁻³ : 1 = 1.36 x 10⁻³

NaI : 0.0625 : 2 = 0.03125

Pb(NO₃)₂ ⇒limiting reactant(smaller ratio)

moles PbI₂ = moles Pb(NO₃)₂ = 1.36 x 10⁻³(mol ratio 1 : 1)

Mass of PbI₂ :

= mol x MW

=  1.36 x 10⁻³ x 461,01 g/mol

= 0.627 g

% yield = 0.55/0.627 x 100% = 87.72%

7 0
3 years ago
Determine a massa, em kg, de um material que está contida em um volume de 18L. Sabe-se que a densidade do material é de 0,9 g/cm
suter [353]
The question in English is "<span>Determine the mass, in kg, of a material that is contained in a volume of 18L. It is known that the material density is 0.9 g/cm 3"

Answer: 
</span>
We can use a simple equation to solve this problem. <span>
    d = m/v</span><span>

<span>Where </span>d <span>is the density, </span>m <span>is the mass and </span>v is the volume.

d = </span>0.9<span> g/cm³
m = ?
v = </span>18 L = 18 x 10³ cm³<span>

By applying the equation,
<span>    0.9 g/cm³ = m / </span></span>18 x 10³ cm³<span>
                  m = 0.9 g/cm³ x </span>18 x 10³ cm³<span> 
<span>                  </span>m = 16200 g
                  m = 16.2 kg

Hence, the mass of 18 L of material is 16.2 kg.</span>
6 0
2 years ago
A student uses a calorimeter to determine the enthalpy of dissolving for ammonium nitrate. The student fills a calorimeter with
ale4655 [162]
Enthalpy change during the dissolution process = m c ΔT,

here, m = total mass = 475 + 125 = 600 g
c = <span>specific heat of water = 4.18 J/g °C
</span>ΔT = 7.8 - 24 = -16.2 oc (negative sign indicates that temp. has decreases)
<span>
Therefore, </span>Enthalpy change during the dissolution = 600 x 4.18 X (-16.2)
                                                                                 = -40630 kJ
(Negative sign indicates that process is endothermic in nature i.e. heat is taken by the system)

Thus, <span>enthalpy of dissolving of the ammonium nitrate is -40630 J/g</span>

7 0
3 years ago
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