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svlad2 [7]
3 years ago
12

A solution of 62.4 g of a covalent compound in enough water to make 1.000 L of solution has an osmotic pressure of 0.305 atm at

25°C. Assume the covalent compound does not dissociate in the solution. Based on these data, what is the molar mass (in g/mol) of covalent? Round your answer to the nearest whole number and do not include units with your answer.
Chemistry
1 answer:
Alborosie3 years ago
8 0

Answer:

4.99 × 10³ g/mol

Explanation:

Step 1: Given  and required data

  • Mass of the covalent compound (m): 62.4 g
  • Volume of the solution (V): 1.000 L
  • Osmotic pressure (π): 0.305 atm
  • Temperature (T): 25°C = 298 K

Step 2: Calculate the molarity (M) of the solution

The osmotic pressure is a colligative pressure. For a covalent compound, it can be calculated using the following expression.

π = M × R × T

M = π / R × T

M = 0.305 atm / (0.0821 atm.L/mol.K) × 298 K

M = 0.0125 M

Step 3: Calculate the moles of solute (n)

We will use the definition of molarity.

M = n / V

n = M × V

n = 0.0125 mol/L × 1.000 L = 0.0125 mol

Step 4: Calculate the molar mass of the compound

0.0125 moles of the compound weigh 62.4 g. The molar mass is:

62.4 g/0.0125 mol = 4.99 × 10³ g/mol

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What charge would an arsenic, As, ion have? Explain.
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Hydrogen sulfide decomposes according to the following reaction: 2H2S(g) ⇋ 2H2(g) + S2(g) Kc=9.30x10-8 at 700.°C.If 0.45 mol of
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Answer:

[H₂] = 1.61x10⁻³ M

Explanation:

2H₂S(g) ⇋ 2H₂(g) + S₂(g)

Kc = 9.30x10⁻⁸ = \frac{[H_{2}]^2[S_{2}]}{[H_{2}S]^2}

First we <u>calculate the initial concentration</u>:

0.45 molH₂S / 3.0L = 0.15 M

The concentrations at equilibrium would be:

[H₂S] = 0.15 - 2x

[H₂] = 2x

[S₂] = x

We <u>put the data in the Kc expression and solve for x</u>:

\frac{(2x^2) * x}{(0.15-2x)^2}=9.30x10^{-8}

\frac{4x^3}{0.0225-4x^2}=9.30*10^{-8}

We make a simplification because x<<< 0.0225:

\frac{4x^3}{0.0225} =9.30*10^{-8}

x = 8.058x10⁻⁴

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A 30.0-g piece of metal at 203oC is dropped into 400.0 g of water at 25.0 oC. The water temperature rises to 29.0 oC. Calculate
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The specific heat of the metal is 2.4733 J/g°C.

Given the following data:

  • Initial temperature of water = 25.0°C
  • Final temperature of water = 29.0°C
  • Mass of water = 400.0 g
  • Mass of metal = 30.0 g
  • Temperature of metal = 203.0°C

We know that the specific heat capacity of water is 4.184 J/g°C.

To find the specific heat of the metal (J/g°C):

Heat lost by metal = Heat gained by water.

Q_{metal} = Q_{water}

Mathematically, heat capacity or quantity of heat is given by the formula;

Q = mc\theta

<u>Where:</u>

  • Q is the heat capacity or quantity of heat.
  • m is the mass of an object.
  • c represents the specific heat capacity.
  • ∅ represents the change in temperature.

Substituting the values into the formula, we have:

30(203)c = 400(4.184)(29 \; -\; 20)\\\\6090c = 1673.6(9)\\\\6090c = 15062.4\\\\c = \frac{15062.4}{6090}

Specific heat capacity of metal, c = 2.4733 J/g°C

Therefore, the specific heat of the metal is 2.4733 J/g°C.

Read more: brainly.com/question/18691577

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