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gulaghasi [49]
3 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]3 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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What types of bonds or interactions are overcome when a nonmetal extended network melts?
STatiana [176]

Covalent bonds or interactions are overcome when a nonmetal extended network melts.

Typically, nonmetals form covalent bonds with one another. A polyatomic ion's atoms are joined by a form of link called covalent bonding. A covalent bond requires two electrons, one from each of the two atoms that are connecting.

One technique to depict the formation of covalent connections between atoms is with Lewis dot formations. The number of unpaired electrons and the number of bonds that can be formed by each element are typically identical. Each element needs to share an unpaired electron in order to establish a covalent bond.

Therefore, covalent bonds or interactions are overcome when a nonmetal extended network melts.

Learn more about covalent bonds here;

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4 0
1 year ago
What is the correct name for this formula: AI2O3?
agasfer [191]

Answer- A

Explanation- Because you have to look at the symbols and remember which symbol goes with each chemical, if you don't then you could get confused.

5 0
3 years ago
Can someone please help me on the second one? I have no idea what to do.
Aleks [24]

The reaction will produce 12.1 g Ag₂S.

<em>Balanced equation</em> = 2Ag + S ⟶ Ag₂S

<em>Mass of Ag₂S</em> = 10.5 g Ag × (1 mol Ag/107.87 g Ag) × (1 mol Ag₂S/2 mol Ag)

× (247.80 g Ag₂S/1 mol Ag₂S) = 12.1 g Ag₂S

8 0
3 years ago
100 POINTS PLEASE HELP!! Honors Stoichiometry Activity Worksheet Instructions: In this laboratory activity, you will taste test
Shtirlitz [24]

Answer:

2 water + sugar + lemon juice → 4 lemonade

Moles of water present in 946.36 g of water=\frac{946.36 g}{236.59 g/mol}=4 mol=

236.59g/mol

946.36g

=4mol

Moles of sugar present in 196.86 g of water=\frac{196.86 g}{225 g/mol}=0.8749 mol=

225g/mol

196.86g

=0.8749mol

Moles of lemon juice present in 193.37 g of water=\frac{193.37 g}{257.83 g/mol}=0.7499 mol=

257.83g/mol

193.37g

=0.7499mol

Moles of lemonade in 2050.25 g of water=\frac{2050.25 g}{719.42 g/mol}=2.8498 mol=

719.42g/mol

2050.25g

=2.8498mol

As we can see that number of moles of lemon juice are limited.

So, we will consider the reaction will complete in accordance with moles of lemon juice.

1 mole lemon juice reacts with 2 mol of water,then 0.7499 mol of lemon juice will react with:

\frac{2}{1}\times 0.7499 mol = 1.4998 mol

1

2

×0.7499mol=1.4998mol of water

Mass of water used = 1.4998 mol × 236.59 g/mol=354.8376 g

Water remained unused = 946.36 g - 354.8376 g =591.5223 g

1 mole lemon juice reacts with mol of sugar,then 0.7499 mol of lemon juice will react with:

\frac{1}{1}\times 0.7499 mol = 0.7499 mol

1

1

×0.7499mol=0.7499mol of water

Mass of sugar used = 0.7499 mol × 225 g/mol = 168.7275 g

Sugar remained unused = 196.86 g - 28.1325 g

1 mole of lemon juice gives 4 moles of lemonade.

Then 0.7499 mol of lemon juice will give:

\frac{4}{1}\times 0.7499 mol=2.996 mol

1

4

×0.7499mol=2.996mol of lemonade

Mass of lemonade obtained = 2.996 mol × 719.42 g/mol = 2157.9722 g

Theoretical yield of lemonade = 2157.9722 g

Experimental yield of lemonade = 2050.25 g

Percentage yield of lemonade:

\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100

theoretical yield

Experimental yield

×100

\frac{2050.25 g}{2157.9722 g}\times 100=95.00\%

2157.9722g

2050.25g

×100=95.00%

6 0
3 years ago
Read 2 more answers
On the pH scale pure water has a pH of<br> A. 0<br> B. 7<br> C. 14
konstantin123 [22]

Answer:its b 7

Explanation:

7 is neutral and pure water is the middle man

7 0
3 years ago
Read 2 more answers
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