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atroni [7]
3 years ago
13

14) For the reaction P4(s) + 6 Hz(9) — 4 PH3(g)

Chemistry
1 answer:
svlad2 [7]3 years ago
3 0
The correct answer is b
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Through an uplift under the earth crust at a dirvergent boundary a _____ is formed.
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<span>Out of the  following given choices;</span>

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6 0
3 years ago
If the container is closed and the ethanol is allowed to reach equilibrium with its vapor, how many grams of liquid ethanol rema
SVETLANKA909090 [29]

Explanation:

Let us assume that the given data is as follows.

        V = 3.10 L,        T = 19^{o}C = (19 + 273)K = 292 K

       P = 40 torr    (1 atm = 760 torr)

So,     P = \frac{40 torr}{760 torr} \times 1 atm

             = 0.053 atm

          n = ?

According to the ideal gas equation, PV = nRT.

Putting the given values into the above equation to calculate the value of n as follows.

                 PV = nRT

   0.053 atm \times 3.10 L = n \times 0.0821 L atm/mol K \times 292 K

                 0.1643 = n \times 23.97

                    n = 6.85 \times 10^{-3}

It is known that molar mass of ethanol is 46 g/mol. Hence, calculate its mass as follows.

               No. of moles = \frac{mass}{\text{molar mass}}

                 6.85 \times 10^{-3} = \frac{mass}{46 g/mol}  

                    mass = 315.1 \times 10^{-3} g

                              = 0.315 g

Thus, we can conclude that the mass of liquid ethanol is 0.315 g.

4 0
3 years ago
(1) The solubility of Salt AB2(S) IS 5mol/dm^3.
Svetlanka [38]

Answer:

a. Ksp = 4s³

b. 5.53 × 10⁴ mol³/dm⁹

Explanation:

a. Obtain an expression for the solubility product of AB2(S),in terms of s.

AB₂ dissociates to give

AB₂ ⇄ A²⁺ + 2B⁻

Since 1 mole of AB₂ gives 1 mole of A and 2 moles of B, we have the mole ratio as

AB₂ ⇄ A²⁺ + 2B⁻

1 : 1 : 2

Since the solubility of AB₂ is s, then the solubility of A is s and that of B is 2s

So, we have

AB₂ ⇄ A²⁺ + 2B⁻

[s]        [s]    [2s]

So, the solubility product Ksp = [A²⁺][B⁻]²

= (s)(2s)²

= s(4s²)

= 4s³

b. Calculate the Ksp of AB₂, given that solubility is 2.4 × 10³ mol/dm³

Given that the solubility of AB is 2.4 × 10³ mol/dm³ and the solubility product Ksp = [A²⁺][B⁻]² = 4s³ where s = solubility of AB = 2.4 × 10³ mol/dm³

Substituting the value of s into the equation, we have

Ksp = 4s³

= 4(2.4 × 10³ mol/dm³)³

= 4(13.824 × 10³ mol³/dm⁹)

= 55.296 × 10³ mol³/dm⁹

= 5.5296 × 10⁴ mol³/dm⁹

≅ 5.53 × 10⁴ mol³/dm⁹

Ksp = 5.53 × 10⁴ mol³/dm⁹

6 0
3 years ago
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