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Katarina [22]
3 years ago
13

(a) what product or products would you expect to obtain by the addition of bromine to cis-cinnamic acid, assuming that it reacts

by the same mechanism as the trans cinnamic acid
Chemistry
1 answer:
Free_Kalibri [48]3 years ago
3 0

If the totaling of bromine to cis-cinnamic acid uses the same mechanism or device as the addition to trans-cinnamic acid, addition of bromine only transpires on the top (syn) and not the bottommost face (anti). Therefore, bromine added to cis-cinnamic only adds to the same face (syn addition) as a result producing only threo dibromides.

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Write the Henderson-Hasselbalch equation for a solution of formic acid. Calculate the quotient [HCO2]/[HCO2H] at (a) pH 3.000; (
Elena L [17]

Answer:

a. 0.182

b. 1.009

c. 1.819

Explanation:

Henderson-Hasselbach equation is:

pH = pKa + log [salt / acid]

Let's replace the formula by the given values.

a. 3 = 3.74 + log [salt / acid]

3 - 3.74 = log [salt / acid]

-0.74 = log [salt / acid]

10⁻⁰'⁷⁴ = 0.182

b. 3.744 = 3.74 + log [salt / acid]

3.744 - 3.74 = log [salt / acid]

0.004 = log [salt / acid]

10⁰'⁰⁰⁴ = 1.009

c. 4 = 3.74 + log [salt / acid]

4 - 3.74 = log [salt / acid]

0.26 = log [salt / acid]

10⁰'²⁶ = 1.819

3 0
3 years ago
PLEASE HELP ME SOLVE THIS.Thank you so much!
Tju [1.3M]

Answer: The coefficients for the given reaction species are 1, 6, 2, 3.

Explanation:

The given reaction equation is as follows.

Cr_{2}O^{2-}_{7} + Cl^{-} \rightarrow Cr^{3+} + Cl_{2}

Now, the two half-reactions can be written as follows.

Reduction half-reaction: Cr_{2}O^{2-}_{7} + 3e^{-} \rightarrow Cr^{3+}

This will be balanced as follows.

Cr_{2}O^{2-}_{7} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O ... (1)

Oxidation half-reaction: Cl^{-} \rightarrow Cl_{2} + 1e^{-}

This will be balanced as follows.

6Cl^{-} \rightarrow 3Cl_{2} + 6e^{-} ... (2)

Adding both equation (1) and (2) we will get the resulting equation as follows.

Cr_{2}O^{2-}_{7} + 14H^{+} + 6Cl^{-} \rightarrow 2Cr^{3+} + 3Cl_{2} + 7H_{2}O

Thus, we can conclude that coefficients for the given reaction species are 1, 6, 2, 3.

6 0
3 years ago
Organisms are classified into different kingdoms based on?...
Naily [24]

Answer:

Food and cell type

Explanation:

Kingdoms are by far the most basic legal structure for living objects. Living objects are classified into realms depending on how they consume food, the kinds of cells that produce their bodies, and the total type of tissue in their bodies.

6 0
3 years ago
A 7.47 g sample of calcium carbonate [CaCO3 (s)] absorbs 85 J of heat, upon which the temperature of the sample increases from 2
Alex_Xolod [135]

Answer:

Specific heat of calcium carbonate(C) = 0.82 (Approx)

Explanation:

Given:

Energy absorbs (q) = 85 J

Change in temperature (Δt) = 34.9 - 21 = 13.9°C  

Mass of calcium carbonate = 7.47 g

Find:

Specific heat of calcium carbonate(C)

Computation:

Specific heat of calcium carbonate(C) = q / m(Δt)

Specific heat of calcium carbonate(C) = 85 / (7.47)(13.9)

Specific heat of calcium carbonate(C) = 85 / 103.833

Specific heat of calcium carbonate(C) = 0.8186

Specific heat of calcium carbonate(C) = 0.82 (Approx)

7 0
3 years ago
As with other ionic compounds, potassium bromate, KBrO3, dissociates into ions when it dissolves in water. If 13.8 g of KBrO3 is
chubhunter [2.5K]

Answer:

ΔH of dissociation is 38,0 kJ/mol

Explanation:

The dissociation reaction of KBrO₃ is:

<em>KBrO₃ → K⁺ + BrO₃⁻ </em>

This dissolution consume heat that is evidenced with the decrease in water temperature.

The heat consumed is:

q = CΔTm

Where C is specific heat of water (4,186 J/mol°C)

ΔT is the temperature changing (18,0°C - 13,0°C = 5,0°C)

And m is mass of water (150,0 mL ≈ 150,0 g)

Replacing, heat consumed is:

q = 3139,5 J ≡ 3,14 kJ

13,8 g of KBrO₃ are:

13,8 g×(1mol/167g) = 0,0826 moles

Thus, ΔH of dissociation is:

3,14kJ / 0,0826mol = <em>38,0 kJ/mol</em>

<em></em>

I hope it helps!

3 0
3 years ago
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