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finlep [7]
3 years ago
10

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

a pressure of 0.970 atm and 390.0 K. What was the initial mass of mercury oxide in the sample?
Chemistry
1 answer:
Yakvenalex [24]3 years ago
7 0

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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Answer:

Lowering the temperature typically reduces the significance of the decrease in entropy. That makes the Gibbs Free energy of the reaction more negative. As a result, the reaction becomes more favorable overall.  

Explanation:

In an addition reaction there's a decrease in the number of particles. Consider the hydrogenation of ethene as an example.

\rm H_2C\text{=}CH_2\; (g) + H_2\; (g) \stackrel{\text{Ni}^\ast}{\to} H_3C\text{-}CH_3\; (g).

When \rm H_2 is added to \rm H_2C\text{=}CH_2 (ethene) under heat and with the presence of a catalyst, \rm H_3C\text{-}CH3 (ethane) would be produced.

Note that on the left-hand side of the equation, there are two gaseous molecules. However, on the right-hand side there's only one gaseous molecule. That's a significant decrease in entropy. In other words, \Delta S < 0.

The equation for the change in Gibbs Free Energy for a particular reaction is:

\Delta G = \Delta H + (\underbrace{- T \, \Delta S}_{\text{entropy}\atop \text{term}}).

For a particular reaction, the more negative \Delta G is, the more spontaneous ("favorable") the reaction would be.

Since typically \Delta S < 0 for addition reactions, the "entropy term" of it would be positive. That's not very helpful if the reaction needs to be favorable.

T (absolute temperature) is always nonnegative. However, lowering the temperature could help bring the value of

8 0
3 years ago
Assuming that the solubility of radon in water with 1 atm pressure of the gas over the water at 30 ∘C is 7.27×10−3M, what is the
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Answer:

K = 137.55 atm/M.

Explanation:

  • The relationship between gas pressure and the  concentration of dissolved gas is given by  Henry’s law:

<em>P = (K)(C)</em>

where P is the partial pressure of the gaseous  solute above the solution (P = 1.0 atm).

k is a constant (Henry’s constant).

C is the concentration of the dissolved gas (C = 7.27 x 10⁻³ M).

∴ K = P/C = (1.0 atm)/(7.27 x 10⁻³ M) = 137.55 atm/M.

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3 years ago
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<span>Two neutral hydrogen atoms share electrons in a covalent bond.
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3 years ago
It is recommended that drinking water contain 1.6 ppm fluo- ride (F) to prevent tooth decay. Consider a cylindrical reservoir wi
Ne4ueva [31]

Answer:

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434  grams of F.

Explanation:

Volume of cylindrical reservoir = V

Radius of the cylindrical reservoir = r = d/2

d = diameter of the cylindrical reservoir = d =4.50\times 10^1 m=45 m

r = d/2 = 22.5 m

Depth of the reservoir = h =  10.0 m

V=\pi r^2 h

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Density of water,d = 1 kg/L

Mass of water cylindrical reservoir =  m

m=d\times V=1 kg/L\times 15,896,250 L=15,896,250 kg

1.6 kilogram of fluorine per million kilograms of water. (Given)

Concentration of fluorine in water = 1.6 kg/ 1000,000 kg of water

In 1000,000 kg of water = 1.6 kg of fluorine

Then 15,896,250 kg of water have x mass of fluorine:

\frac{x}{15,896,250 kg\text{kg of water}}=\frac{1.6 kg}{1000,000 \text{kg of water}}

x=\frac{1.6 kg}{1000,000}\times 15,896,250 kg=25.434 kg

15,896,250 kg water of contains mass 25.434 kg of fluorine.

25.434 kg = 25434 g

25,434  grams of fluorine  should be added to give 1.6 ppm.

Percentage of fluorine in hydrogen hexafluorosilicate :

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m' = 32,127.02 g

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434  grams of F.

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