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Stels [109]
3 years ago
9

What is state symbols

Chemistry
1 answer:
Ganezh [65]3 years ago
5 0

Answer:

Explanation:

State symbols are used in chemical equations to delineate the state of matter in which the reaction is taking place.

They give a good perspective of the state of the reactants and products obtainable.

There are basically four states of matter in every chemical reaction:

  • Solids are symbolized by small letter (s)
  • Liquids are represented by (l)
  • Gases are shown by (g)
  • Aqueous solutions having water as the medium by (aq)

These symbols appear as subscript in front of the chemical species.

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The normal freezing point of a certain liquid
slavikrds [6]

Answer : The molal freezing point depression constant of liquid X is, 4.12^oC/m

Explanation :  Given,

Mass of urea (solute) = 5.90 g

Mass of liquid X (solvent) = 450 g  = 0.450 kg

Molar mass of urea = 60 g/mole

Formula used :  

\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}}{\text{Molar mass of urea}\times \text{Mass of liquid X Kg}}

where,

\Delta T_f = change in freezing point

\Delta T_s = freezing point of solution = -0.5^oC

\Delta T^o = freezing point of liquid X = 0.4^oC

i = Van't Hoff factor = 1 (for non-electrolyte)

K_f = Molal-freezing-point-depression constant = ?

m = molality

Now put all the given values in this formula, we get

0.4^oC-(-0.5^oC)=1\times K_f\times \frac{5.90g}{60g/mol\times 0.450kg}

K_f=4.12^oC/m

Therefore, the molal freezing point depression constant of liquid X is, 4.12^oC/m

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The student placed 10 mL of PbCl2 (saturated solution) in the test tube and added a pinch of NaCl. A white precipitate of PbCl2
NNADVOKAT [17]
PbCl₂(aq) → Pb²⁺(aq) + 2Cl⁻(aq)

NaCl(aq) → Na⁺(aq) + Cl⁻(aq)

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A concentrated aqueous solution of Pb(NO3)2 is slowly added to 1.0 L of a mixed aqueous solution containing 0.010 M Na2CrO4 and
Oduvanchick [21]

Answer:

This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M

Explanation:

Step 1: Data given

Molarity of Na2CrO4 = 0.010 M

Molarity of NaBr = 2.5 M

Ksp(PbCrO4) = 1.8 * 10^–14

Ksp(PbBr2) = 6.3 * 10^–6

Step 2: The balanced equation

PbCrO4 →Pb^2+ + CrO4^2-

PbBr2  → Pb^2+ + 2Br-

Step 3: Define Ksp

Ksp PbCrO4 = [Pb^2+]*[CrO4^2-]

1.8*10^-14 = [Pb^2+] * 0.010 M

[Pb^2+] = 1.8*10^-14 /0.010

[Pb^2+] = 1.8*10^-12 M

The minimum [Pb^2+] needed to precipitate PbCrO4 is 1.8*10^-12 M

Ksp PbBr2 = [Pb^2+][Br-]²

6.3 * 10^–6 = [Pb^2+] (2.5)²

[Pb^2+] = 1*10^-6 M

The minimum [Pb^2+] needed to precipitate PbBr2 is 1*10^-6 M

This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M

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2 years ago
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