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Sidana [21]
3 years ago
6

Why should the water in the calorimeter be stirred throughout the reaction?

Chemistry
1 answer:
Readme [11.4K]3 years ago
7 0
I think this is done to allow even distribution of temperature in all the regions of the calorimeter. That is to make sure that there are no hot or cold spots and also ensure that the measured temperature is the actual average temperature when the readings are taken by a thermometer. 
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Determine the percentage of carbon and hydrogen in ethane C2H6 if the molecular weight is 30.
Naddika [18.5K]

Answer:

Percentage of carbon:

{ \tt{ =  \frac{24}{30}  \times 100\%}} \\  = 80\%

Percentage of hydrogen:

{ \tt{ =  \frac{6}{30}  \times 100\%} } \\  = 20\%

8 0
3 years ago
You are the medical examiner at the scene of a murder at a restaurant. There is a
Hoochie [10]

The compound : C₄₀H₄₄N₄O

<h3>Further explanation</h3>

The empirical formula is the smallest comparison of atoms of compound =mole ratio of the components

The principle of determining empirical formula

• Determine the mass ratio of the constituent elements of the compound.  

• Determine the mole ratio by dividing the percentage by the atomic mass

The mol ratio of composition : C : H : N : O

\tt \dfrac{80.66}{12}\div \dfrac{7.39}{1}\div \dfrac{9.39}{14}\div \dfrac{2.68}{16}\\\\6.722\div 7.39\div 0.671\div 0.1675\rightarrow divide~by~smallest(0.1675)\\\\40\div 44\div 4\div 1

3 0
3 years ago
Is it possible for the equivalence point of a titration to not be at pH 7? Explain your answer.
lara31 [8.8K]
<span>The reason it will be 7 for some titrations is that when you  titrates a strong acid with a strong base for example  HCl and NaOH the salt formed is conjugate base of strong acid and will be a very weak base
 That means that it cannot produce any OH^-1 and all the H+ has been converted to water.The only source of H+ or OH is water with a Ka of 10^-14 so the pH = -log [H+]=-log 10^-7 = 7 
second reason is 
When you titrates a weak acid with strong base at equivalence point 
only a water solution of the conjugate base exists 

CH3COOH + NaOH ----- Na+ CH3COO^-1 + H2O 
Since the conjugate base is the conjugate base of a weak acid it will hydrolyze in water like so 
for instance Na+ CH3COO^-1 + HCl---- CH3COOH + NaCl the equivalence point will be way BELOW 7 and in the case of above will be less than 5. So pH of 7 at equivalence point is only reached in strong acid strong base titrations.
hope this helps</span>
5 0
3 years ago
Read 2 more answers
Which of the following is the balanced equation for P + O2 → P2O5? Question 3 options: A) 4P + O2 → 2P2O B) P + 5O2 → 2P2O5 C) 4
Luden [163]

Given equation:

P + O2 → P2O5

In order for the equation to be balanced, the stoichiometry of the atoms of one kind on the reactant side must be equal to that on the product

Reactants                                            Products

P = 1                                                       P = 2

O = 2                                                     O = 5

The balanced equation would be:

4P + 5O2 → 2P2O5

Reactants                                            Products

P = 4                                                      P = 4

O = 10                                                    O = 10

Ans: D)


4 0
3 years ago
Read 2 more answers
Construct a three-step synthesis of trans-2-pentene from acetylene by dragging the appropriate formulas into the bins. Note that
adelina 88 [10]

Answer:

The three-step synthesis of trans-2-pentene from acetylene is as follows.

<u>Step -1:</u> Formation of higher order terminal alkyne on reaction with sodium acetylides with haloalkanes.

<u>Step -2:</u> Formation terminal alkyne to nonterminal alkynes.

<u>Step -3:</u> Formation of trans-pent - 2-pent-ene by reduction.

Explanation:

Synthesis of trans-pent-2-yne from ethyne takes place is mainly a three step synthesis which involves formation of higher order terminal alkyne on reaction with sodium acetylides with haloalkane. Second step involves the further alkylation of terminal alkynes to higher order nonterminal alkynes and the third step involves the formation of trans-2-ene by dissolving reduction method.

The chemical reaction of each step of chemical reactions is as follows.

8 0
3 years ago
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