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Rom4ik [11]
3 years ago
8

At a certain temperature, the solubility of strontium arsenate, sr3(aso4)2, is 0.0480 g/l. what is the ksp of this salt at this

temperature?
Chemistry
1 answer:
Nadusha1986 [10]3 years ago
8 0
Given the solubility of strontium arsenate is 0.0480 g/l . we have to convert it into mol/L by dividing it over molar mass (540.7 g/mol)
Molar solubility = 0.0480 / 540.7 = 8.9 x 10⁻⁵ mol/L
Dissociation equation:
Sr₃(AsO₄)₂(s) → 3 Sr²⁺(aq) + 2 AsO₄³⁻(aq)
                             3 s                   2 s
Ksp = [Sr²⁺]³ [AsO₄³⁻]²
       = (3s)³ (2s)²
       = 108 s⁵
Ksp = 108 (8.9 x 10⁻⁵) = 5.95 x 10⁻¹⁹ 
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The volume of  0. 250 mole sample of H_{2} gas occupy if it had a pressure of 1. 70 atm and a temperature of 35 °C is  3.71 L.

Calculation,

According to ideal gas equation which is known as ideal gas law,

PV =n RT

  • P is the pressure of the hydrogen gas  = 1.7 atm
  • Vis the volume of the hydrogen gas = ?
  • n is the number of the hydrogen gas = 0.25 mole
  • R is the universal gas constant = 0.082 atm L/mole K
  • T is the temperature of the sample = 35°C = 35 + 273 = 308 K

By putting all the values of the given data like pressure temperature universal gas constant and number of moles in equation (i) we get ,

1.7 atm×V = 0.25 mole ×0.082 × 208 K

V = 0.25 mole ×0.082atm L /mole K × 308 K /1.7 atm

V = 3.71 L

So, volume of the sample of the hydrogen gas occupy is  3.71 L.

learn more about ideal gas equation

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Before you touch an electrical switch, plug, or outlet
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Which of the following occurs as temperature increases?
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A particular radioactive nuclide has a half-life of 1000 years. What percentage of an initial population of this nuclide has dec
Delicious77 [7]

Answer:

91.16% has decayed & 8.84% remains

Explanation:

A = A₀e⁻ᵏᵗ => ln(A/A₀) = ln(e⁻ᵏᵗ) => lnA - lnA₀ = -kt => lnA = lnA₀ - kt

Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹

Time (t) = 1000yrs  

A = fraction of nuclide remaining after 1000yrs

A₀ = original amount of nuclide = 1.00 (= 100%)  

lnA = lnA₀ - kt

lnA = ln(1) – (6.93 x 10ˉ⁴yrsˉ¹)(3500yrs) = -2.426

A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years

Amount of nuclide decayed = 1 – 0.0884 = 0.9116 or 91.16% has decayed.

3 0
3 years ago
If you have 730 moles of a gas at 304 Kelvin in a 4.50 Liter container what
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Answer:

4046atm

Explanation:

For this question you can use the ideal gas law,

<em />PV=nRT<em />

Where P is pressure, V is volume, n is moles of substance, R is the constant, and T is the temperature.

Because of the units given, R will equal .08026P=4046 atm

<h3>Rearrange the equation to solve for pressure:</h3>

P=\frac{nRT}{V}

Then, plug in the values (I'll be excluding units for simplicity, but they all cancel out for pressure in atm):

P=\frac{(730)(.08206)(304)}{4.5}

This will give you:

P=4046atm

3 0
3 years ago
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