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Gemiola [76]
2 years ago
7

Be sure to answer all parts. provide stepwise mechanisms illustrating how each product is formed by choosing, in order, the step

s involved in the formation of each product. if a mechanistic step is not needed, select "n/a". qs "b:" = lda and/or −oc(ch3)3 a a b b c c 06 d e e 08 f 07 g h h i i 09] j k k i. step [1] step [2] p1 step [3] ii. step [1] step [2] p2 step [3] iii. step [1] step [2] p3 step [3] iv. step [1] step [2] p4

Chemistry
1 answer:
SpyIntel [72]2 years ago
8 0

Answer:

Explanation:

find the solution below

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Imagine you needed to identify if an object has undergone a physical change or a chemical change. What information would you nee
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Much to everyone’s surprise, nitrogen monoxide (NO) has been found to act as a neurotransmitter. To prepare to study its effect,
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Answer: 0.0043mole

Explanation:Please see attachment for explanation

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2 years ago
"The elementary reaction 2 NO2(g) → 2 NO(g) + O2(g) is second order in NO2 and the rate constant at 600 K is 6.77 × 10-1 M-1s-1.
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Answer : The half-life at this temperature is, 3.28 s

Explanation :

To calculate the half-life for second order the expression will be:

t_{1/2}=\frac{1}{k\times [A_o]}

When,

t_{1/2} = half-life = ?

[A_o] = initial concentration = 0.45 M

k = rate constant = 6.77\times 10^{-1}M^{-1}s^{-1}

Now put all the given values in the above formula, we get:

t_{1/2}=\frac{1}{6.77\times 10^{-1}M^{-1}s^{-1}\times 0.45M}

t_{1/2}=3.28s

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7 0
2 years ago
A chemist combined chloroform (CHCl3) and acetone (C3H6O) to create a solution where the mole fraction of chloroform is 0.187. T
ladessa [460]

Answer:

\large \boxed{\text{c = 2.50 mol/L; b = 3.96 mol/kg }}

Explanation:

1. Molar concentration

Let's call chloroform C and acetone A.

Molar concentration of C = Moles of C/Litres of solution

(a) Moles of C

Assume 0.187 mol of C.

That takes care of that.

(b) Litres of solution

Then we have 0.813 mol of A.

(i) Mass of each component

\text{Mass of C} = \text{0.187 mol C} \times \dfrac{\text{119.38 g C}}{\text{1 mol C}} = \text{22.32 g C}\\\\\text{Mass of A} = \text{0.813 mol A} \times \dfrac{\text{58.08 g A}}{\text{1 mol A}} = \text{47.22 g A}

(ii) Volume of each component

\text{Vol. of C} = \text{22.32 g C} \times \dfrac{\text{1 mL C}}{\text{1.48 g C}} = \text{15.08 mL C}\\\\\text{Vol. of A} = \text{47.22 g A} \times \dfrac{\text{1 mL A}}{\text{0.791 g A}} = \text{59.70 mL A}

(iii) Volume of solution

If there is no change of volume on mixing.

V = 15.08 mL + 59.70 mL = 74.78 mL

(c) Molar concentration of C

c = \dfrac{\text{0.187 mol}}{\text{0.07478 L}} = \textbf{2.50 mol/L }\\\\\text{ The molar concentration of chloroform is $\large \boxed{\textbf{2.50 mol/L}}$}

2. Molal concentration of C

Molal concentration = moles of solute/kilograms of solvent

Moles of C = 0.187 mol

Mass of A = 47.22 g = 0.047 22 kg

\text{b} = \dfrac{\text{0.187 mol}}{\text{0.047 22 kg}} = \textbf{3.96 mol/kg }\\\\\text{The molal concentration of chloroform is $\large \boxed{\textbf{3.96 mol/kg}}$}

4 0
2 years ago
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