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Oksanka [162]
3 years ago
11

Water (H2O) forms when hydrogen gas (H2) and oxygen gas (O2) react according to the equation below: 2H2 O2 → 2H2O What mass of w

ater forms when 1. 45 × 10-3 g O2 react completely? (Molar mass of O2 = 32. 00 g/mol; molar mass of H2O = 18. 02 g/mol) 1. 63 × 10-3 g 8. 16 × 10-4 g 1. 29 × 10-3 g.
Chemistry
1 answer:
AveGali [126]3 years ago
8 0

The mass of water formed in the reaction is \rm 1.63\;\times\;10^-^3 g. Thus, option A is correct.

The balanced chemical equation for the formation of water is:

\rm 2\;H_2\;+\;O_2\;\rightarrow\;2\;H_2O

From the balanced chemical equation, 1 mole of oxygen forms 2 moles of water.

<h3>Computation for the mass of water</h3>

The moles of a compound is given by:

\rm Moles=\dfrac{Mass}{Molar\;mass}

The moles of oxygen reacted in the reaction are:

\rm Moles\;O_2=\dfrac{1.45\;\times\;10^-^3\;g}{32\;g/mol} \\Moles\;O_2=0.045\;\times\;10^-^3\;mol

The moles of water formed is:

\rm 1\;mol\;O_2=2\;mol\;H_2O\\0.045\;\times\;10^-^3\;mol\;O_2=0.045\;\times\;10^-^3\;\times\;2\;mol\;H_2O\\0.045\;\times\;10^-^3\;mol\;O_2=0.09\;\times\;10^-^3\;mol\;H_2O

The moles of water formed are \rm 0.09\;\times\;10^-^3 mol.

The mass of water is given as:

\rm Mass=Moles\;\times\;Molar\;mass\\Mass\;H_2O=0.09\;\times\;10^-^3\;\times\;18\;g\\Mass\;H_2O=1.63\;\times\;10^-^3\;g

The mass of water formed in the reaction is \rm 1.63\;\times\;10^-^3 g. Thus, option A is correct.

Learn more about mass formed, here:

brainly.com/question/7324123

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Explanation:

a) The solubility (S) is the concentration of the salt that will be dissociated and form the ions in the solution, the solubility product constant (Kps) is the multiplication of the concentration of the ions elevated at their coefficients. The concentration of the ions depends on the stoichiometry and will be equivalent to S.

The salts solubilization reactions and their Kps values are:

MA(s) ⇄ M⁺²(aq) + A⁻²(aq) Kps = S*S = S²

MZ₂(aq) ⇄ M⁺²(aq) + 2Z⁻(aq) Kps = S*S² = S³

Thus, the Kps of MZ₂ has a larger value.

b) A saturated solution is a solution that has the maximum amount of salt dissolved, so, the concentration dissolved is solubility. As we can notice from the reactions, the concentration of M⁺² is the same for both salts.

c) The equilibrium will be not modified because the salts have the same solubility. So, let's suppose that the volume of each one is 1 L, so the number of moles of the cation in each one is 4x10⁻⁴ mol. The total number of moles is 8x10⁻⁴ mol, and the concentration is:

8x10⁻⁴ mol/2 L = 4x10⁻⁴ mol/L.

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This is an exercise in<u> the General Combined Gas Law</u>.

To start solving this exercise, we obtain the following data:

<h3>Data:</h3>
  • V₁ = 4.00 l
  • P₁ = 365 mmHg
  • T₁ = 20 °C + 273 = 293 K
  • V₂ = 2,80 l
  • T₂ = 30 °C + 273 = 303 K
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We apply the following formula:

  • P₁V₁T₂=P₂V₂T₁     ⇒  General formula

Where:

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We clear for final pressure (P2)

\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{P_{1}V_{1}T_{2}}{V_{2}T_{1}} \ \ \to \ \ \ Formula \end{gathered}$}

We substitute our data into the formula:

\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{(365 \ mmHg)(4.00 \not{l})(303 \not{K})}{(2.80 \not{l})(293\not{K})}  \end{gathered}$}

\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{442380 \ mmHg}{ 820.4 }  \end{gathered}$}

\boxed{\large\displaystyle\text{$\begin{gathered}\sf P_{2}=539.224 \ mmHg \end{gathered}$}}

Answer: The new canister pressure is 539.224 mmHg.

<h2>{ Pisces04 }</h2>
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