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vesna_86 [32]
3 years ago
14

You have two sealed jars of water at the same temperature. in the first jar there is a large amount of water. in the second jar

there is a small amount of water. using 3 -4 sentences explain how the vapor pressure of water in the first jar compares with the vapor pressure of water in the second jar.
Chemistry
1 answer:
n200080 [17]3 years ago
4 0
The vapor pressure of the jar with a large amount of water and the jar with a small amount of water is the same. Vapor pressure is an intrinsic property. This means that this property is not dependent on the amount of the substance. Vapor pressure is dependent only on temperature and since the temperature is the same for both jars, their vapor pressures are also the same.
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Write the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by each:
Maru [420]

Answer :

(a) The charge and full ground-state electron configuration of the monatomic ion is, (+1) and 1s^22s^22p^63s^23p^64s^23d^{10}4p^6

(b) The charge and full ground-state electron configuration of the monatomic ion is, (-3) and 1s^22s^22p^6

(c) The charge and full ground-state electron configuration of the monatomic ion is, (-1) and 1s^22s^22p^63s^23p^64s^23d^{10}4p^6

Explanation :

For the neutral atom, the number of protons and electrons are equal. But, they are unequal when the atoms present in the form of ions or the atom has some charges.

When an unequal number of electrons and protons then it leads to the formation of ionic species.

Ion : An ion is formed when an atom looses or gains electron.

When an atom looses electrons, it will form a positive ion known as cation.

When an atom gains electrons, it will form a negative ion known as anion.

(a) The given element is, Rb (Rubidium)

As we know that the rubidium element belongs to group 1 and the atomic number is, 37

The ground-state electron configuration of Rb is:

1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^1

This element will easily loose 1 electron and form Rb^+ ion  which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Rb ion is:

1s^22s^22p^63s^23p^64s^23d^{10}4p^6

(b) The given element is, N (Nitrogen)

As we know that the nitrogen element belongs to group 15 and the atomic number is, 7

The ground-state electron configuration of N is:

1s^22s^22p^3

This element will easily gain 3 electrons and form N^{3-} ion  which attain stable noble gas electronic configuration.

The full ground-state electron configuration of N ion is:

1s^22s^22p^6

(c) The given element is, Br (Bromine)

As we know that the bromine element belongs to group 17 and the atomic number is, 35

The ground-state electron configuration of Rb is:

1s^22s^22p^63s^23p^64s^23d^{10}4p^5

This element will easily gain 1 electron and form Br^- ion  which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Br ion is:

1s^22s^22p^63s^23p^64s^23d^{10}4p^6

4 0
3 years ago
Is dew forming on a leaf a physical or chemical change?
solong [7]
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5 0
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Which events represents a chemical change ?
8_murik_8 [283]
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Calculate the amount of grams of water that was used when 5525J of energy was used to boil this amount of water.
djverab [1.8K]

Answer:

2.445 g

Explanation:

Step 1: Given and required data

  • Energy in the form of heat required to boil the water (Q): 5525 J
  • Latent heat of vaporization of water (∆H°vap): 2260 J/g
  • Mass of water (m): ?

Step 2: Calculate the mass of water

We will use the following expression.

Q = ∆H°vap × m

m = Q / ∆H°vap

m = 5525 J / (2260 J/g)

m = 2.445 g

4 0
2 years ago
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