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Flura [38]
2 years ago
12

The molar mass of oxygen gas (O2) is 32.00 g/mol. The molar mass of C3H8 is 44.1 g/mol. What mass of O2, in grams, is required t

o completely react with 0.025 g C3H8
Chemistry
1 answer:
Ira Lisetskai [31]2 years ago
6 0
The reaction formula of this is C3H8 + 5O2 --> 3CO2 + 4H2O. The ratio of mole number of C3H8 and O2 is 1:5. 0.025g equals to 0.025/44.1=0.00057 mole. So the mass of O2 is 0.00057*5*32=0.0912 g.
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Identify the precipitate(s) formed when solutions of na2so4(aq), ba(no3)2(aq), and nh4clo4(aq) are mixed.
Lostsunrise [7]

Answer : BaSO_{4} will be the precipitate which will be formed.


Explanation : When all the three solutions namely; NaSO_{4}  + Ba(NO_{3})_{2}  + NH_{4} ClO_{4} are mixed together a white precipitate of BaSO_{4} is formed as a product in the solution along with the soluble by product of Ammonium nitrate which is NH_{4} NO_{3} 

7 0
2 years ago
Zinc has a mass of 4.50g calculate the volume
Neko [114]
Divide the mass by the density

4 0
2 years ago
As noted in Assignment 1.4.1., alkylation of benzene with 1-chlorobutane in the presence of AlCl3 gives both butylbenzene and (1
Sunny_sXe [5.5K]

Answer:

Use the Bromotriflouride catalyst, BF₃

Explanation:

The BF₃ is most likely to yield less desired side products. The effect lies in the reaction mechanism.

BF₃ is a Lewis acid. Its role is to promote the ionization of the HF. This is achieved through the electrophilic mechanism. The reaction mechanism is as follows:

2 - methylpropene + H-F-BF₃ → H-F + H₃C + benzene

butylbenzene + F-BF₃ → tert-butylbenzene + H-F + BF₃ (regenerated catalyst)

6 0
2 years ago
Use the following half-reactions to construct a voltaic cell:
velikii [3]

<u>Answer:</u> The correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

<u>Explanation:</u>

We are given:

Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction: Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V

Reduction half reaction: Ag^{+}+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V       ( × 3)

Net equation:  3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)

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}=0.80-(-0.73)=1.53V

Hence, the correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

4 0
2 years ago
3. The following data of decomposition reaction of thionyl chloride (SO2Cl2) were collected at a certain temperature and the con
KonstantinChe [14]

Answer:

a) First-order.

b) 0.013 min⁻¹

c) 53.3 min.

d) 0.0142M

Explanation:

Hello,

In this case, on the attached document, we can notice the corresponding plot for each possible order of reaction. Thus, we should remember that in zeroth-order we plot the concentration of the reactant (SO2Cl2 ) versus the time, in first-order the natural logarithm of the concentration of the reactant (SO2Cl2 ) versus the time and in second-order reactions the inverse of the concentration of the reactant (SO2Cl2 ) versus the time.

a) In such a way, we realize the best fit is exhibited by the first-order model which shows a straight line (R=1) which has a slope of -0.0013 and an intercept of -2.3025 (natural logarithm of 0.1 which corresponds to the initial concentration). Therefore, the reaction has a first-order kinetics.

b) Since the slope is -0.0013 (take two random values), the rate constant is 0.013 min⁻¹:

m=\frac{ln(0.0768)-ln(0.0876)}{200min-100min} =-0.0013min^{-1}

c) Half life for first-order kinetics is computed by:

t_{1/2}=\frac{ln(2)}{k}=\frac{ln(2)}{0.013min^{-1}}  =53.3min

d) Here, we compute the concentration via the integrated rate law once 1500 minutes have passed:

C=C_0exp(-kt)=0.1Mexp(-0.013min^{-1}*1500min)\\\\C=0.0142M

Best regards.

6 0
2 years ago
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