I need help with this. Read instructions and help plzz

a = Proton transfer d = SN2 Nucleophilic substitution g = Nucleophilic subs at carbonyl(acyl Xfer) b = Lewis acid/base e= Electr

marissa [1.9K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The solution to this question is shown on the second uploaded image

Explanation:

4 0

3 years ago

(First to answer gets some good) Order the following components of the universe from largest to smallest: Moon, planet,universe,

cluponka [151]

Answer:

Universe, galaxy, solar system, star, planet, moon and asteroid.

Explanation:

You're welcome!

7 0

3 years ago

When 20.0 g of KI are dissolved in 50.0 mL of distilled water in a calorimeter, the temperature drops from 24.0 °C to 19.0 °C. C

Reil [10]

<h2>Answer:</h2>

<h2>Explanations</h2>

The formula for calculating the amount of heat absorbed by the water is given as:

$\begin{gathered} q=mc\triangle t \\ q=50\times4.18\frac{J}{g^oC}\times(19-24) \\ q=50\times4.18\times(-5) \\ q=-1045Joules \\ q=-1.045kJ \end{gathered}$

Determine the moles of KI

$\begin{gathered} moles\text{ of KI}=\frac{mass\text{ of KI}}{molar\text{ mass of KI}} \\ moles\text{ of KI}=\frac{20g}{166g\text{/mol}} \\ moles\text{ of KI}=0.1205moles \end{gathered}$

Since heat is lost, hence the enthalpy change of the solution will be negative that is:

$\begin{gathered} \triangle H=-q \\ \triangle H=-(-1.045kJ) \\ \triangle H=1.045kJ \end{gathered}$

Determine the enthalpy of solution in kJ•mol-1

$\begin{gathered} \triangle H_{diss}=\frac{1.045kJ}{0.1205mole} \\ \triangle H_{diss}\approx8.67kJmol^{-1} \end{gathered}$

Hence the enthalpy of solution in kJ•mol-1 for KI is 8.67kJ/mol

4 0

1 year ago

PLEASE HELP ME The pH scale for acids and bases ranges from _____.

inysia [295]

The pH scale for acids and bases ranges from 1 - 14. The answer is letter C. The rest of the choices do not answered the question above. There are quite a few relationships between [H+] and [OH−] ions. And because there is a large range of number between 10 to 10-15 M, the pH is used. pH = -log[H+] and pOH = -log[OH−]. In aqueous solutions, [H+ ][OH- ] = 10-14. From here we can derive the values of each concentration.

-log[H+ ] + -log[OH- ] = -log[10-14]

pH + pOH = 14

So pH = 14 – pOH and pOH = 14 –

pH

It would be B.

7 0

3 years ago

()

const2013 [10]

Answer:

$\large \boxed{\text{-1276 kJ/mol}}$

Explanation:

You calculate the energy required to break all the bonds in the reactants.

Then you subtract the energy needed to break all the bonds in the products.

CH₃CH₂OH + 3O₂ ⟶ 2CO₂ + 3H₂O

Bonds: 5C-H 1C-C 1C-O 1O-H 3O=O 4C=O 6O-H

D/kJ·mol⁻¹: 413 347 358 467 495 799 467

$\Delta H = \sum{D_{\text{reactants}}} - \sum{D_{\text{products}}}\\\sum{D_{\text{reactants}}} = 5 \times 413 + 1 \times 347 + 1 \times 358 + 1 \times 467 + 3 \times 495 = 3237 + 1485\\=\text{4722 kJ}\\\sum{D_{\text{products}}} = 4 \times 799 + 6 \times 467 =3196 + 2802 = \text{5998 kJ}\\\Delta H = 4722 - 5998= \textbf{-1276 kJ} \\ \text{The overall energy change is $\large \boxed{\textbf{-1276 kJ/mol}}$}.$

6 0

3 years ago