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mixer [17]
3 years ago
7

Draw a resonance structure, complete with all formal charges and lone (unshared) electron pairs, that shows the resonance intera

ction of the acetoxy with the ortho position in phenyl acetate.
Chemistry
1 answer:
Angelina_Jolie [31]3 years ago
8 0

Answer:

See explanation

Explanation:

The acetoxy group in ortho position in phenyl acetate does interact with the phenyl moiety in the molecule via resonance.

This detailed interaction of the  acetoxy group in ortho position in phenyl acetate with the phenyl moiety in the molecule via resonance is shown in the image attached.

This interaction is made possible because the oxygen atom of the acetoxy group has lone pairs of electrons that are suitably positioned to interact with the ring via resonance.

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Creativity plays a vital role in science. In what aspects of science is creativity involved? A. Designing experiments B. Formula
Citrus2011 [14]

Answer:

Option D. All of these

Explanation:

The creativity involves the  process of using the brain to use new or existing knowledge to overcome a problem. This sometimes involves starting something purely from stretch or using the existing design laws and theories to solve a problem. As a result, the correct option is D as all of the above applies.  

4 0
3 years ago
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What is the formula for the compound copper (II) sulfate pentahydrateA. C2S3•6H2O
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Answer:

D

copper (II) sulfate pentahydrate

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CuSO _{4}.5H _{2}O

8 0
3 years ago
If he(g) has an average kinetic energy of 8750 j/mol under certain conditions, what is the root mean square speed of cl2(g) mole
saw5 [17]
<span>The root mean square speed is given by V_rms = âšRT/M where r, t, and m are the rate constant, temperature and molar mass the gas Average molar kinetic energy of the gas E = 1/2 M * (V_rms)^2 = 8750 ms/1 So (V_rms)^2 = (2 * 8750) / M Molar mass of 2 chlorine atoms in kg is 2 * 35 * 10^(-3) Hence we have (V_rms)^2 = (2 * 8750)/ (2 * 35 * 10^(-3)) (V_rms)^2 = 8750/0.035 = 250000 So V_rms = âš 250000 = 500</span>
5 0
3 years ago
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A nuclear reactor core must stay at or below 95 °C to remain in good working condition. Cool water at a temperature of 10 °C is
aliina [53]

Answer:

\large \boxed{\text{67 000 g}}

Explanation:

This is a problem in calorimetry — the measurement of the quantities of heat that flow from one object to another.

It is based on the Law of Conservation of Energy — Energy can be transformed from one type to another, but it cannot be destroyed or created.

If heat flows out of the reactor (negative), the same amount of heat must flow into the water (positive).

Since there is no change in total energy,

heat₁ + heat₂ = 0

The symbol for the quantity of heat transferred is q, so we can rewrite the word equation as

q₁ + q₂  = 0

The formula for the heat absorbed or released by an object is

 q = mCΔT, where

 m = the mass of the sample

  C = the specific heat capacity of the sample, and

ΔT = T_f - T_i = the change in temperature

1. Equation

There are two heat flows in this problem,

heat released by reactor + heat absorbed by water = 0

               q₁                  +                        q₂                     = 0

               q₁                  +                 m₂C₂ΔT₂                 = 0

2. Data:

q₁ = -23 746 kJ

m₂ = ?; C₂ = 4.184 J°C⁻¹g⁻¹;  T_f = 95 °C; T_i = 10 °C

3. Calculations

(a) Convert kilojoules to joules

q_{1} = -\text{23 746 kJ} \times \dfrac{\text{1000 J}}{\text{1 kJ}} = -\text{23 746 000 J}

(b) ΔT  

ΔT₂ = T_f - T_i = 95 °C - 10 °C = 85 °C

(c) m₂

\begin{array}{rcl}q_{1} + q_{2} & = & 0\\\text{-23 746 000 J} + m_{2} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 85 \, ^{\circ}\text{C} & = & 0\\\text{-23 746 000 J} + 356m_{2} \text{J$\cdot$g}^{-1} & = & 0\\356m_{2} \text{g}^{-1} & = & 23746000\\m_2&=& \dfrac{23746000}{\text{356 g}^{-1}}\\\\ & = & \textbf{67000 g}\\\end{array}\\

\text{You must circulate $\large \boxed{\textbf{67 000 g}}$ of water each hour.}

7 0
3 years ago
What is the empirical formula for a compound if a sample contains 3.72 g of P and 21.28 g of Cl
fenix001 [56]
M(P)=3.72 g
M(P)=31 g/mol

m(Cl)=21.28 g
M(Cl)=35.5 g/mol

n(P)=m(P)/M(P)
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n(Cl)=m(Cl)/M(Cl)
n(Cl)=21.28/35.5=0.60 mol

P : Cl = 0.12 : 0.60 = 1 : 5

PCl₅ - is the empirical formula
6 0
3 years ago
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