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tangare [24]
3 years ago
9

Eueueeueueueuehuehuehuehuehuehueuehuehuehuek

Chemistry
1 answer:
zlopas [31]3 years ago
4 0

Answer:

Please don not post  Questions that don'y make sense

Explanation:

P.s: Thanks for the point and have anice day

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Balance the chemical equation. Based on the equation, how many grams of bromine are produced by the complete reaction of 11 gram
scoundrel [369]

Answer:

Explanation:

The balanced chemical equation is Cl2 + 2KBr → 2KCl + Br2. The amount of bromine is calculated as follows:

11.0 g KBr (1mol KBr/119.002g KBr * 1mol Br2/2mol KBr * 159.808g Br2/1mol Br2= 7.39 g Br2.)

5 0
3 years ago
Write a mechanism for the esterification of propanoic acid with 18O-labeled ethanol. Show clearly the fate of the 18O label. (b)
tatuchka [14]

Answer:

See explanation and images attached

Explanation:

a) In the mechanism for the acid catalysed esterification of propanoic acid using ethanol, we can see that the first step is the protonation of the acid followed by nucleophillic attack of the alcohol. Loss of water and consequent deprotonation regenerates the acid catalyst. We can see the fate of the 18O labelled ethanol in the mechanism shown.

b)  In the second mechanism, an unnamed ester is hydrolysed using an acid catalyst. The attack of the acid and subsequent nucleophillic attack of water labelled with 18O leads to the incorporation of this 18O into the product acid as shown in the mechanism attached to this answer.

5 0
3 years ago
Does all Newton's laws work together
9966 [12]
I believe so ya.....
6 0
3 years ago
A thin-walled sphere rolls along the floor. What is the ratio of its translational kinetic energy to its rotational kinetic ener
ICE Princess25 [194]

Explanation:

The kinetic energy of translation

E_1=\frac{1}{2}mv^2

m= mass v= linear velocity

The kinetic energy of rotation

E_2=\frac{1}{2}I\omega^2

I= MOI of the thin walled sphere =kmR^2

where ω= v/R= angular velocity

E_2=\frac{1}{2}kmR^2\frac{v}{R}^2

Then

\frac{E_1}{E_2} = \frac{0.5mv^2}{0.5kmR^2\frac{v}{R}^2 }

=1/k

solid sphere: k=0.4;   E1/E2 =1/0.4 = 2.5;  

 hollow sphere: k=2/3;   E1/E2 = 1.5

3 0
3 years ago
What is the free energy change if the ratio of the concentrations of the products to the concentrations of the reactants is 22.3
snow_lady [41]

The free energy change(Gibbs free energy-ΔG)=-8.698 kJ/mol

<h3>Further explanation</h3>

Given

Ratio of the concentrations of the products to the concentrations of the reactants is 22.3

Temperature = 37 C = 310 K

ΔG°=-16.7 kJ/mol

Required

the free energy change

Solution

Ratio of the concentration : equilbrium constant = K = 22.3

We can use Gibbs free energy :

ΔG = ΔG°+ RT ln K

R=8.314 .10⁻³ kJ/mol K

\tt \Delta G=-16.7~kJ/mol+8.314.10^{-3}\times 310\times ln~22.3\\\\\Delta G=-8.698~kJ/mol

8 0
4 years ago
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