As the speed of the particles decreases, -a. Intermolecular forces become stronger

b. Intermolecular forces become weaker

c. Intermolecular forces do not change

d. Energy increases
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Answers

Answer 1
Answer:

As the speed of the particles decreases, intermolecular forces become stronger. Thus, the correct option for this question is A.

What are Intermolecular forces?

The intermolecular forces may be defined as the forces of attraction. that is present between atoms, molecules, and ions when they are placed close to each other in order to form a compound or element. This force is continuously acting on the neighboring particles of different molecules.

It is found that at low temperature, when the speed of molecules/particles decrease, they migrate closer to one another that results in their intermolecular forces that become stronger as compared to the initial one.

As the attraction between molecules gradually increases, their movement decreases, and undergo fewer collisions between them.

Therefore, as the speed of the particles decreases, intermolecular forces become stronger. Thus, the correct option for this question is A.

To learn more about Intermolecular forces, refer to the link:

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Answer 2
Answer:

Answer:

B

Explanation:


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Enter the oxidation number of one atom of each element in each reactant and product.CH4(g)+2O2(g)⟶CO2(g)+2H2O(g)
C in CH4 :
H in CH4 :
O in O2 :
C in CO2 :
O in CO2 :
H in H2O :
O in H2O :
Which atom is reduced?
Which atom is oxidized?

Answers

The oxidation numbers of the atoms of the specified elements in each of the given atoms are;

1) -4

1) -42) +1

1) -42) +13) 0

1) -42) +13) 04) +4

1) -42) +13) 04) +45) -2

1) -42) +13) 04) +45) -26) +1

1) -42) +13) 04) +45) -26) +17) -2

1) -42) +13) 04) +45) -26) +17) -2Atom oxidized = C

1) -42) +13) 04) +45) -26) +17) -2Atom oxidized = CAtom reduced = O

1) C in CH4

To get the oxidation number of C;

Oxidation state of hydrogen atom is +1 and so if the oxidation state of C is x, then we have;

x + 4(+1) = 0

x + 4 = 0

x = -4

2) H in CH4

Oxidation state on Carbon atom in this case is -4. Thus;

-4 + 4x = 0

4x = 4

x = +1

3) O in O2

This is oxygen gas that exists in it's free state and as such oxidation number is 0.

4) C in CO2

Oxidation state of O here is -2. Thus;

x + 2(-2) = 0

x - 4 = 0

x = +4

5) O in CO2

Oxidation state of C is +4 here. Thus;

4 + 2x = 0

2x = -4

x = -4/2

x = -2

6) H in H2O

Oxidation state of oxygen here is -2. Thus;

2x - 2 = 0

2x = 2

x = 2/2

x = +1

7) O in H2O

Oxidation state of hydrogen here is +1. Thus;

2(1) + x = 0

x = -2

Finally, oxidation number of carbon increased, then it is the atom that was oxidized while the atom reduced is the Oxygen atom.

Read more at; brainly.com/question/22816291

Answer:

1. -4

2. +1

3. 0

4. +4

5. -2

6. +1

7. -2

reduced = H

oxidized = O

Explanation:

Know oxidation rules.

- Hope this helped! Please let me know if you would like to learn this. I could show you the rules and help you work through them.

1. An isotope of cesium-137 has a half-life of 30 years. If 5.0 g of cesium-137 decays over 60 years, how many grams will remain?

Answers

Answer:

1.25 g

Explanation:

Now we have to use the formula;

N/No = (1/2)^t/t1/2

N= mass of cesium-137 left after a time t (the unknown)

No= mass of cesium-137 present at the beginning = 5.0 g

t= time taken for 5.0 g of cesium-137 to decay =60 years

t1/2= half life of cesium-137= 30 years

Substituting values;

N/5= (1/2)^60/30

N/5= (1/2)^2

N/5= 1/4

4N= 5

N= 5/4

N= 1.25 g

Therefore, 1.25 g of cesium-137 will remain after 60 years.

Which enzymes show very broad substrate specificity and can therefore be used to produce several higher alcohols.Select one:

a. 2-keto acid decarboxylase AND alcohol dehydrogenase
b. alcohol dehydrogenase
c. transaminase
d. acetolactate synthasee. 2-ketoacid decarboxylase

Answers

Answer:

A. 2-keto acid decarboxylase and alcohol dehydrogenase

Explanation:

2-keto acid decarboxylase and alcohol dehydrogenase are used to produce many higher alcohols. These enzymes also display a high degree of specificity on their substrate

Which element is a halogen?chlorine (Cl)
oxygen (O)
carbon (C)
radon (Rn)

Answers

The element that has been a halogen from the given elements have been Chlorine. Thus, option A is correct.

Halogen has been given as the group of metallic compounds that have been consisted of the one valence electrons in their last orbitals.

The halogens have been the most reactive elements and belong to group 18 of the periodic table. From the given set of elements, the number of valence electrons in the following has been:

  • Chlorine = 1 valence electron
  • Oxygen = 6 valence electrons
  • Carbon = 4 valence electrons
  • Radon = 8 valence electrons

Since the element having 1 valence electron has been Chlorine. Thus, the element that has been a halogen from the given elements have been Chlorine. Thus, option A is correct.

For more information about halogen, refer to the link:

brainly.com/question/2288388

Answer is chlorine (Cl)

Draw the Lewis structure for the polyatomic hydronium H3O cation. Be sure to include all resonance structures that

Answers

Answer:

 Lewis structure of Hydronium ion is shown below :                          

Explanation:

Lewis structure : It is a representation of valence electrons on the atoms in a molecule

Here , Hydronium ion is given , which contains 1 atom of oxygen and 3 atoms of hydrogen .

Oxygen has a total of 6 valence electrons and hydrogen contains 1 valence electron .

Oxygen share its 3 valence electrons with 3 hydrogen atoms and left with 3 valence electrons. From these three valence  electrons of oxygen atom  two electrons will be shown as a pair of electrons on oxygen atom but a single electron can not be shown . So , to simplify this, one positive charge is shown overall .  

Resonance structure will be same as the hybrid structure because all  three atoms are same , that is hydrogen .

How many atoms does 6H2O contain

Answers

Answer:

two atoms of oxygen. For H2O, there is one atom of oxygen and two atoms of hydrogen.