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Sergio [31]
3 years ago
8

What is the chemical reaction taking place?

Chemistry
1 answer:
ale4655 [162]3 years ago
7 0

Answer: Synthesis (of phosphorous and oxygen to form tetraphosphorous decaoxide)

Explanation:

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In a certain chemical process, water reacts with a compound, and the water is converted into another compound during the reactio
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Answer:

hydration reaction

Explanation:

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CH_2=CH_2(g)   +   H_2O(g)  ⇄  CH_3CH_2OH(g)

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How did Albert Einstein explain the photoelectric effect?
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3. During an experiment where 50.0 mL of a 1.0 M acid solution was mixed with 50.0 mL of a 1.0 M base solution, the temperature
Aleksandr-060686 [28]

Answer:

\large \boxed{\text{-61 kJ$\cdot$mol$^{-1}$}}  

Explanation:

Data:

                H₃O⁺ +  OH⁻ ⟶ 2H₂O

    V/mL:  50.0     50.0

c/mol·L⁻¹:   1.0        1.0

     

    ΔT = 6.5 °C  

      ρ = 1.210 g/mL

      C = 4.18 J·°C⁻¹g⁻¹

C_cal = 12.0 J·°C⁻¹

Calculations:

(a) Moles of acid

\text{Moles of acid} = \text{0.0500 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{0.0500 mol}\\\\\text{Moles of base} = \text{0.0500 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{0.0500 mol}

(b) Volume of solution

V = 50.0 mL + 50.0 mL = 100.0 mL

(c) Mass of solution

\text{Mass of solution} = \text{100.0 mL} \times \dfrac{\text{1.10 g}}{\text{1 mL}} = \text{110.0 g}

(d) Calorimetry

There are three energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm the water

q₃ = heat to warm the calorimeter

         q₁      +           q₂         +       q₃      = 0

     nΔH      +       mCΔT      + C_calΔT = 0

0.0500ΔH + 1.10×4.18×6.5 + 12.0×6.5 = 0

0.0500ΔH +        2989       + 78.0       = 0

                             0.0500ΔH + 3067 = 0

                                          0.0500ΔH = -3067

                                                      ΔH = -3067/0.0500

                                                            = -61 000 J/mol

                                                            = -61 kJ/mol

\text{The enthalpy of reaction is $\large \boxed{\textbf{-61 kJ$\cdot$mol$^{\mathbf{-1}}$}}$}

Note: The answer can have only two significant figures because that is all you gave for the change in temperature.

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