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3 years ago

2A1Cl3 + 3Na CO3→ Al2(CO3)3 + 6NaCl

Chemistry

1 answer:

forsale [732]3 years ago

4 0

Answer:

1. aluminum chloride

2. 3

3. 9

Explanation:

1. Know the names

2. reactants are on left side products are on right side

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The enthalpy of formation of MX is ΔHf° = –525 kJ/mol. The enthalpy of sublimation of M is ΔHsub = 139 kJ/mol. The ionization en

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2 years ago

Determine the direction of bond polarity for each compound. Leave it blank for a nonpolar bond. Br−Br H−Cl F−CH3

NeX [460]

Answer:

For H-Cl, the direction is towards the chlorine atom

For F-CH3, the direction is towards the flourine atom.

Explanation:

The dipole moment is a vector quantity. This implies that it has both magnitude and direction.

Thus, the direction of the dipole moment always points from the positive atom towards the negative atom.

This explains the fact that it points to chlorine in HCl and points to flourine in F-CH3

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3 years ago

You are making a 0.2M solution of sodium acetate in the lab. Sodium acetate has a molecular mass of 82.03 g/mol. How many g of s

lyudmila [28]

Answer: The answer can be found on CHEG

Explanation:

7 0

3 years ago

The molar solubility of pbi2 is 1.5 103 m.

Vsevolod [243]

Answer: -

Concentration of PbI₂ = 1.5 x 10⁻³ M

PbI₂ dissociates in water as

PbI₂ ⇄ Pb²⁺ + 2 I⁻

So PbI₂ releases two times the amount of I⁻ as it's own concentration when saturated.

Thus the molar concentration of iodide ion in a saturated PbI₂ solution = [ I⁻] =

= 1.5 x 10⁻³ x 2 M

= 3 x 10⁻³ M

PbI₂ releases the same amount of Pb²⁺ as it's own concentration when saturated.

[Pb²⁺] = 1.5 x 10⁻³ M

So solubility product for PbI₂

Ksp = [Pb²⁺] x [ I⁻]²

=1.5 x 10⁻³ x (3 x 10⁻³)²

= 4.5 x 10⁻⁹

8 0

2 years ago

What is the standard enthalpy of a reaction for which the equilibrium constant is (a) doubled, (b) halved when the temperature i

Alexxandr [17]

Answer:

a) 48KJ

b) -48KJ

Explanation:

Given that;

ln(K2/K1) = ΔH°/R(1/T2 - 1/T1)

K2= equilibrium constant at T2

K1 = equilibrum constant at T1

R = gas constant

T1 = initial temperature

T2 = final temperature

When we double the equilibrium constant K1; K2 = 2K1

T1 = 310 K

T2 = 310 + 15 = 325 K

ln(2K1/K1) =- ΔH°/R(1/T2 - 1/T1)

ln2 = -ΔH°/8.314(1/325 - 1/310)

0.693 = -ΔH°/8.314(3.08 * 10^-3 - 3.2 * 10^-3)

0.693 = -ΔH°/8.314 (-0.00012)

0.693 = 0.00012ΔH°/8.314

0.693 * 8.314 = 0.00012ΔH°

ΔH° = 0.693 * 8.314/0.00012

ΔH° = 48KJ

b) K2 =K1/2

ln(K1/2/K1) =- ΔH°/R(1/T2 - 1/T1)

ln (1/2) = -ΔH°/8.314 (1/325 - 1/310)

-0.693 = -ΔH°/8.314 (-0.00012)

-0.693 = 0.00012ΔH°/8.314

-0.693 * 8.314 = 0.00012ΔH°

ΔH°= -0.693 * 8.314/0.00012

ΔH°= -48KJ

6 0

3 years ago