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Aleks04 [339]
3 years ago
13

The amount of energy needed to change a material from a solid to a liquid is heat of

Chemistry
1 answer:
GalinKa [24]3 years ago
7 0
It revolves around Physical change.  I learned this in class today.

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A when sulfuric acid is added to sirie, hydrager
irga5000 [103]

Sulfuric acid is prepared industrially by means of the response of water with sulfur trioxide which in turn is made by way of a chemical combination of sulfur dioxide and oxygen both by way of the touch system or the chamber system.

                         H2SO4 (l) H2O (g) + SO3 (g).

The reaction is highly exothermic as an enormous amount of heat is liberated.

The usual approach is to dilute the sulfur trioxide in sulphuric acid. This produces oleum. SO3 (g) + H2SO4 → H2S2O7 (1) Oleum can be in addition diluted in water to acquire concentrated sulphuric acid.

An acid catalyst is added to protonate the carbonyl carbon. How does this catalyze the response, robust acid catalysts catalyze the hydrolysis and transesterification of esters which enables the mechanism with a view to boom the electrophilicity of the carbonyl carbon to assist protonate the carbonyl oxygen.

Learn more about sulfuric acid here:-brainly.com/question/11857977

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8 0
2 years ago
Is the air that we breathe an element, compound, mixture, or pure substance?
ddd [48]
The answer is mixture
6 0
3 years ago
Which is the first stage of mitosis? Group of answer choices Anaphase Prophase Metaphase Telophase
ss7ja [257]

Answer:Prophase is the first stage of mitosis. In prophase,

Explanation:

3 0
3 years ago
Read 2 more answers
A galvanic cell at a temperature of 42 degrees Celcius is powered by the following redox reaction:
seropon [69]

<u>Answer:</u> The cell voltage of the given reaction is 1.86 V

<u>Explanation:</u>

The given chemical equation follows:

3Cu^{2+}(aq.)+2Al\rightarrow 2Al^{3+}(aq.)+2Au(s)

<u>Oxidation half reaction:</u> Al(aq.)\rightarrow Al^{3+}(aq.)+3e^-;E^o_{Al^{3+}/Al}=1.66V       ( × 2)

<u>Reduction half reaction:</u> Cu^{2+}(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.16V       ( × 3)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.16-(-1.66)=1.82V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Al^{3+}]^2}{[Cu^{2+}]^3}

where,

E_{cell} = electrode potential of the cell = ? V

E^o_{cell} = standard electrode potential of the cell = +1.82 V

n = number of electrons exchanged = 2

R = Gas constant = 8.314 J/mol Kl

T = temperature = 42^oC=[42+273]K=315K

F = Faraday's constant = 96500

[Al^{3+}]=1.63M

[Cu^{2+}]=3.43M

Putting values in above equation, we get:

E_{cell}=1.82-\frac{2.303\times 8.314\times 315}{2\times 96500}\times \log(\frac{(1.63)^2}{(3.43)^3})\\\\E_{cell}=1.86V

Hence, the cell voltage of the given reaction is 1.86 V

4 0
3 years ago
Which of the following can lower the activation energy of a system?
aleksandrvk [35]

wrong boiiiiiiiiiiii

8 0
4 years ago
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