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gulaghasi [49]
4 years ago
13

suggest one reason why 1g of cobalt oxide nanoparticles is a better catalyst than 1g of cobalt oxide powder

Chemistry
1 answer:
anastassius [24]4 years ago
7 0
Particles are smaller, a reactant would be exposed to more cobalt atoms. Causing the reaction to happen quicker than with larger particles.
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A rock that weighs 12.2 g is added to a graduated cylinder with 30.0 mL of water in it. The meniscus rises to 54.4 mL. What is t
GuDViN [60]

Answer:

              0.50 g/mL

Explanation:

Formula of Density,

                          D = mass / volume

Mass = 12.2 g

Volume = 54.4 mL - 30.0 mL = 24.4 mL

Putting values,

                          D = 12.2 g / 24.4 mL

                          D = 0.50 g/mL

4 0
2 years ago
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6 0
3 years ago
Question 7-Which of the following is not an acid-base neutralization reaction?
docker41 [41]

Answer:- It is choice D. H_2SO_4+2NaCl\rightarrow 2HCl+Na_2SO_4

Explanations:- In general, a neutralization reaction is the reaction of an acid with base to form salt and water.

in first reaction the reaction is taking place between ammonia(a base) and hydrochloric acid(an acid) to form their salt(ammonium chloride). So, it is an acid-base neutralization reaction.

In second reaction, sodium hydroxide(a base) is reacting with acetic acid(an acid) to form their salt(sodium acetate) and water. So, it is an acid-base neutralization reaction.

In third reaction, Nitric acid is reacting with calcium hydroxide(a base) to form a salt(calcium nitrate) and water. So, it is an acid-base neutralization reaction.

In fourth reaction, sulfuruc acid is reacting a sodium chloride(a salt) to give a double replacement reaction. It is not an acid-base neutralization reaction as it's not taking place between an acid and base.

So, the correct choice is D. H_2SO_4+2NaCl\rightarrow 2HCl+Na_2SO_4

7 0
3 years ago
Read 2 more answers
Consider the conversion of 2-naphthol and 1-bromobutane into an ether via the williamson ether synthesis. list the procedural st
Lemur [1.5K]

Answer:

Following are the solution to this question:

Explanation:

Please find the complete question in the attachment.

Start of Laboratory

Dissolve 2-naphthol in the round bottom flask with ethanol.

Add pellets of sodium hydroxide and hot chips. Attach a condenser.

Heat for 20 minutes under reflux, until the put a burden dissolves.

After an additional hour, add 1-Bromobutane and reflux.

Pour the contents into a beaker with ice from a round bottom flask.

On a Bachner funnel, absorb the supernatant by vacuum filtration.

Utilizing cold water to rinse the material and dry that on the filter.

Ending of the Lab

6 0
3 years ago
Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir
guajiro [1.7K]

Answer:

y_{O2} =4.3%

Explanation:

The ethanol combustion reaction is:

C_{2}H_{5} OH+3O_{2}→2CO_{2}+3H_{2}O

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}

Dividing the previous equation by x:

1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles

Calculate the number of moles of CO2 and water considering the same:

0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)

0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)

The total number of moles at the reactor output would be:

N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)

So, the oxygen mole fraction would be:

y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3%

6 0
3 years ago
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