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IrinaK [193]
3 years ago
5

1. Use the equation below for the following question.

Chemistry
1 answer:
zhuklara [117]3 years ago
3 0

Answer:

120.44 × 10²³ molecules

Explanation:

Given data:

Moles of C₃H₈ = 5.0 mol

Molecules of water produced = ?

Solution:

Chemical equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Now we will compare the moles of C₃H₈ and water.

                     C₃H₈           :         H₂O

                         1              ;             4

                        5              :            4×5 = 20 mol

Molecules of water:

1 mole = 6.022 × 10²³ molecules

20 mol×6.022 × 10²³ molecules / 1 mol

120.44 × 10²³ molecules

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Answer:

Hi... Your answer is 10*50=500

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2 years ago
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The osmotic pressure of a saturated solution of strontium sulfate at 25 ∘C∘C is 21 torrtorr. Part A What is the solubility produ
USPshnik [31]

Answer

solubility product = 3.18x 10^-7

Explanation:

We were given the pressure in torr then we need to convert to atm for consistency, ten we have

21torr/760= 0.0276315789 atm

21 Torr = .0276315789 atm

P = i M S T

M = P / iRT

Where p is osmotic pressure

T= temperature= 25C+ 273= 298K

for XY vanthoff factor i = 2

S = 0.0821 L-atm / mol K

M = .0276315789 atm / (2)(0.0821 L atm / K mole)(298 K)

M = 0.000564698046 mol/liters

solubility= 0.000564698046 mol/liters

Ksp = [X+][Y-]

Ksp = X^2

Ksp = [Sr^+2] * [SO4^-2]

Ksp = X^2

Ksp = (0.000564698046)^2

Ksp = 3.18883883 × 10-7

Ksp = 3.18x 10^-7

solubility product = 3.18x 10^-7

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7 0
3 years ago
An unknown amount of mercury (II) oxide was decomposed in the lab. Mercury metal was formed and 5.20 L of oxygen was released at
Yakvenalex [24]

Answer:

  • <u>68.3g</u>

Explanation:

<u>1. Word equation:</u>

  • <em>mercury(II) oxide → mercury + oxygen </em>

<u>2. Balanced molecular equation:</u>

  • 2HgO → 2Hg + O₂(g)

<u>3. Mole ratio</u>

Write the ratio of the coefficients of the substances that are object of the problem:

       2molHgO/1molO_2

<u>4. Calculate the number of moles of O₂(g)</u>

Use the equation for ideal gases:

          pV=nRT\\\\\\n=\dfrac{pV}{RT}\\\\\\n=\dfrac{0.970atm\times5.20L}{0.08206atm.L/K.mol\times 390.0K}\\\\\\n=0.1576mol

<u>5. Calculate the number of moles of HgO</u>

         \dfrac{2molHgO}{1molO_2}\times 0.1576molO_2=0.315molHgO

<u>6. Convert to mass</u>

  • mass = # moles × molar mass

  • molar mass of HgO: 216.591g/mol

  • mass = 0.315mol × 216.591g/mol = 68.3g

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