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aleksandr82 [10.1K]
3 years ago
10

At a certain temperature, the solubility of strontium arsenate, Sr3(AsO4)2, is 0.0540 g/L. What is the Ksp of this salt at this

temperature
Chemistry
1 answer:
blondinia [14]3 years ago
7 0
 <span>Get into moles first. .0590 grams over 540.8 grams per mole = 1.09 x l0^-4 moles 

Sr3(As04)2 = 3 Sr++(aq) plus 2 As04^-3(aq) 

Ksp = (Sr++)^3(As04^-3)^2 

(Sr++) = 3 X l.09 x l0^-4 = 3.27 x l0^-4 

(As04^-3) = 2 x l.09 x l0^-4 = 2.18 x l0^-4 

<span>Ksp = (3.27 x l0^-4)^3 (2.18 x l0^-4)^2 which equals 1.66 x 10^-18th


I hope my answer has come to your help. Thank you for posting your question here in Brainly.</span></span>
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Zanzabum

<em>K</em> = 5.0 × 10^25

<h2>Part (a). Calculate <em>E</em>° for the reaction </h2>

<em>Step 1.</em> Write the equations for the two half-reactions

2H^(+)(aq) + 2e^(-) → H2(g); _0.00 V

Zn^(2+)(aq) + 2e^(-) → Zn(s); -0.76 V

<em>Step 2.</em> Identify the cathode and the anode

The half-cell with the more negative <em>E</em>° (Zn) is the anode.

<em>Step 3.</em> Calculate <em>E</em>°

Zn(s) → Zn^(2+)(aq) + 2e^(-); _________+0.76 V

2H^(+)(aq) + 2e^(-) → H2(g); __________0.00 V

Zn(s) + 2H^(+)(aq) → Zn^(2+)(aq) + H2(g); +0.76 V

<em>E</em>° = +0.76 V

<h2>Part (b). Calculate <em>K</em> for the reaction </h2>

The relation between <em>E</em>° and <em>K</em> is

<em>E</em>° = (<em>RT</em>)/(<em>nF</em>)ln<em>K </em>

where

<em>R</em> = the universal gas constant: 8.314 J·K^(-1)mol^(-1)

<em>T</em> = the Kelvin temperature

<em>n</em> = the moles of electrons transferred

<em>F</em> = the Faraday constant: 96 485 J·V^(-1)mol^(-1)

Then

0.76 V = [8.314 J·K^(-1)mol^(-1) × 298.15 K]/[2 × 96 485 J·V^(-1)mol^(-1)]ln<em>K</em>

0.76 = 0.012 85 ln<em>K</em>

ln<em>K</em> = 0.76/0.012 85 = 59.16

<em>K</em> =e^59.16 = 5.0 × 10^25

4 0
3 years ago
Which answer best describes what is happening in the following reaction?
NISA [10]

A.  This is not a redox reaction. It is an example of combustion.

<h3>Combustion reaction of hydrocarbon</h3>

During the combustion of a hydrocarbon, the hydrocarbon reacts with oxygen to create carbon dioxide, water, and heat.

<h3>Example of combustion reaction</h3>

2C8H18 + 25O2 → 16CO2 + 18H2O

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1 year ago
Is C2S6 ionic or covalent
Stels [109]

Answer:

covalent

Explanation:

C2S6 is a compound made of 2 nonmetals which makes it covalent

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8 0
3 years ago
Lead−206 is the end product of 238u decay. one 206pb atom has a mass of 205.974440 amu. (a) calculate the binding energy per nuc
Dovator [93]
The atomic number for Pb is 82
∴ Pb has 82 protons and 206-82 = 14 protons
The actual mass of Pb nuclei is
=(82 × mass of the proton) + (124 × mass of neutron)
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= 207.6713 amu
The mass of lead which is given is 205.9744 amu
∴mass defect is
m = 207.6713 - 205.9744 = 1.6969 amu
=1.6969 × 1.66054 × 10⁻²⁷kg
=2.818 × 10⁻²⁷kg
The binding energy is E = mc²
C is the speed of light in vacuum = 2.9979 × 10⁸m/s
∴ E = 2.532 × 10×⁻¹⁰ J/mol
= 2.532 × 10⁻¹⁰ × 6.023 × 10²³ J/mol 
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8 0
3 years ago
A 20g piece of lead absorbs 566 joules of heat and its temperature changes from 35 oC to 195 oC. Calculate the specific heat.
Alja [10]

Answer:

  • <u>Question 1: 0.2J/(gºC)</u>
  • <u>Question 2: 6,000J</u>
  • <u>Question 3: 300J</u>
  • <u>Question 4: 80g</u>
  • <u>Question 5: 74ºC</u>
  • <u>Question 6: 50g</u>

<u></u>

Explanation:

Question 1.<em> A 20g piece of lead absorbs 566 joules of heat and its temperature changes from 35º oC to 195º C. Calculate the specific heat.</em>

<em />

The thermal energy equation is:

  • Q = m × C × ΔT

<em />

Substitute and solve for C:

  • 566J = 20g × C × (195ºC - 35ºC)
  • C = 566J / (20g × 160ºC)
  • C = 0.177 J/(gºC) ≈ 0.2J/(gºC)

<em />

You must round to one significant figure because one factor has one significant figure).

<em />

<em />

Qustion 2.<em> 40g of water is heat at 40ºC and the temperature rise to 75ºC. What is the amount of heat needed for the temperature to rise? (specific heat of water is 4.184 J/gºC)</em>

<em />

Use the thermal energy equation again:

  • Q = m × C × ΔT

<em />

Substitute and compute:

  • Q = 40g × 4.184 J/gºC × (75ºC - 40ºC)
  • Q = 5,857.6J

Round to one significant figure: 6,000J

<em />

Question 3. <em>Graphite has a mass of 50g and a specific heat of 0.420 J/gºC. If graphite is cooled from 50ºC to 35ºC, how much energy was lost?</em>

  • Q = m × C × ΔT
  • Q = 50g × 0.420J/gºC × (35ºC - 50ºC)
  • Q = 315J

Round to one significant figure (because 50g has one significant figure)

  • Q = 300J

<em />

Question 4.<em> </em><em>Iron has a specific heat of 0.712 J/gºC. A piece of iron absorbs 3000J of energy and undergoes a temperature change totaling 50ºC, What is the mass of iron?</em>

<em />

  • Q = m × C × ΔT

Solve for m:

  • m = Q / (C × ΔT)

Substitute and compute:

  • m = 3,000J / (0.712J/gºC × 50ºC)
  • m = 84.26 g ≈ 80 g (rounded to one significant figure, because the factor 3,000J has one significant figure).

Question 5. <em>If 400g of an unknown solution at 70ºC loses 7500 J of heat, what is the final temperature of the unknown solution. The unknown solution has a specific heat of 4.184 J/gºC.</em>

<em />

  • Q = m × C × ΔT

<em />

Q is negative, since it is released.

Substitute and solve for T:

  • - 7,500J = 400g × 4.184J/gºC × (T - 70ºC)

  • T = - 7500J / 400g × 4.184J/gºC) + 70ºC

  • T = 74ºC

<em />

If you round to one significant figure you cannot tell the temperature difference, thus leave two significant figures.

<em />

Question 6. <em>How many grams of water would require 9500J of heat to raise the temperature from 50ºC to 100ºC</em>

  • Q = m × C × ΔT

Subsitute:

  • 9,500J = m × 4.184J/gºC × (100ºC - 50ºC)

Solve for m and compute:

  • m = 9,500J / (4.184J/gºC × 50ºC)

  • m = 45g

Since the temperatures indicate one singificant figure, the mass should be rounded to one significant figure:

  • m = 50g.
8 0
3 years ago
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