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ioda
3 years ago
14

Select all the true statements. When an atom gains an electron, it becomes a cation. Anions carry a positive charge. The Cl− and

Br− ions have the same number of electrons. The K+ ion is formed when a potassium atom loses one electron. The Fe2+ and Fe3+ ions have the same number of protons. The Cu+ and Cu2+ ions have the same number of electrons.
Chemistry
1 answer:
REY [17]3 years ago
3 0

Answer:

The statements 4 and 5 are true.

Explanation:

1. When an atom gains an electron it becomes negatively charged. This negatively charged species is called anion.

A + e⁻ →  A⁻ (anion)

Therefore, the statement 1 is false.

2.  An anion is formed when an atom gains an electron and becomes negatively charged. Therefore, an anion is a negatively charged species.

A + e⁻ →  A⁻ (anion)

Therefore, the statement 2 is false.

3. The atomic number of chlorine atom Cl is 17 and atomic number of bromine atom Br is 35.

Since, for neutral atom, the atomic number of an atom is equal to the number of electrons present in that atom.

Therefore, the number of electrons in Cl atom is 17 and the number of electrons in Br atom is 35.

When the Cl atom gains one electron it forms Cl⁻ ion and when the Br atom gains one electron it forms Br⁻ ion.

Therefore, the number of electrons in Cl⁻ ion is 17 + 1 = 18 electrons

and the number of electrons in Br⁻ ion is 35 + 1 = 36 electrons

Therefore, Cl⁻ and Br⁻ ions do not have the same number of electrons.

Therefore, the statement 3 is false.

4. When potassium atom (K) loses one electron it forms a positively charged species called potassium cation (K⁺).

K  → K⁺ + e⁻

Therefore, the statement 4 is true.

5. The atomic number of Fe atom is 26.

Since, the atomic number of an atom is equal to the number of protons present in that atom.

When the Fe atom loses two electrons to form Fe²⁺ and when the Fe atom loses three electrons to form Fe³⁺ ion, the number of protons remains the same.

Therefore, the ions Fe²⁺ and Fe³⁺ have the same number of protons.

Therefore, the statement 5 is true.

6. The atomic number of copper atom Cu is 29.

Since, for neutral atom, the atomic number of an atom is equal to the number of electrons present in that atom.

Therefore, the number of electrons in Cu atom is 29

When the Cu atom loses one electron it forms Cu⁺ ion and when the Cu atom loses two electrons it forms Cu²⁺ ion.

Cu  → Cu⁺ + e⁻                and    Cu  → Cu²⁺ + 2e⁻

Therefore, the number of electrons in Cu⁺ ion is 29 - 1 = 28 electrons

and the number of electrons in Cu²⁺ ion is 29 - 2 = 27 electrons

Therefore,  Cu⁺ ion and Cu²⁺ ion do not have the same number of electrons.

Therefore, the statement 6 is false.

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ira [324]

Explanation:

The given data is as follows.

        m = 10.0 kg = 10,000 g    (as 1 kg = 1000 g)

      Initial temp. of block 1, T_{1} = 100^{o}C = (100 + 273) K = 373 K  

      Initial temp. of block 2, T_{2} = 0^{o}C = (0 + 273) K = 273 K

So, heat released by block 1 = heat gained by block 2

            mC \Delta T = mC \times \Delta T

  10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C

                  T_{f} - 100^{o}C = 0^{o}C - T_{f}    

                   2T_{f} = 100^{o}C

                          T_{f} = 50^{o}C

Convert temperature into kelvin as (50 + 273) K = 323 K.              

Also, we know that the relation between enthalpy and temperature change is as follows.

             \Delta H = mC \Delta T

                         = 10000 g \times 0.385 J/K g \times 323 K

                         = 1243550 J

or,                     = 1243.5 kJ

Now, calculate entropy change for block 1 as follows.

     \Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}

            = 10000 g \times 0.385 J/K g \times ln \frac{323}{373}

            = 10000 g \times 0.385 J/K g \times -0.143

            = -554.12 J/K

Now, entropy change for block 2 is as follows.

   \Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}

           = 10000 g \times 0.385 J/K g \times ln \frac{323}{273}

           = 10000 g \times 0.385 J/K g \times 0.168

           = 647.49 J/K

Hence, total entropy will be sum of entropy change of both the blocks.

            \Delta S_{total} = \Delta S_{1} + \Delta S_{2}

                       = -554.12 J/K + 647.49 J/K

                       = 93.37 J/K

Thus, we can conclude that for the given reaction \Delta H is 1243.5 kJ and \Delta S_{total} is 93.37 J/K.

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KATRIN_1 [288]

Answer:

gain or loses electrons.

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Which molecule will undergo only london dispersion forces when interacting with other molecules of the same kind?.
Neko [114]

Molecules undergo London dispersion forces:

C_{4}H_{10} is the molecule will undergo only London dispersion forces when interacting with other molecules of the same kind.

What are London dispersion forces?

  • A sort of force that interacts between atoms and molecules that is often electrically symmetric is referred to as a London dispersion force.
  • When viewed from the nucleus, their electron distribution is frequently symmetrical. This dispersion force, which is also known as a transient attractive force, is frequently observed when the locations of the electrons in two nearby atoms cause the atoms to temporarily form dipoles.
  • The bond is polar when there are significant variations between the elements' electronegativities; it is nonpolar when there are similarities. When the molecule's dipole moment is equal to O, it is nonpolar; when it differs from O, it is polar.
  • The force at these molecules is known as the London dispersion force. In nonpolar molecules, the forces are weak, and partial charges must be induced so that they can bond. In polar molecules, partial charges caused by polarity result in a stronger link known as a dipole-dipole. The dipole-dipole is significantly stronger and known as a hydrogen bond if it is connected to a large electronegative atom (F, O, or N). Ionic force is the name for the attraction force at ionic substances.
  • The intermolecular force in the letter an is the London dispersion force because the compound is nonpolar;

<u>Reason for incorrect options:</u>

b: the compound is ionic because Na is a metal and the other part is covalent,

c: two compounds are possible: one is nonpolar and exhibits London dispersion force; the other is polar and exhibits dipole-dipole force; and

d: both compounds exhibit hydrogen bonds (H bonded to O, and H bonded to F).

NOTE: Your question is incomplete, but most probably your full question was, which molecule will undergo only London dispersion forces when interacting with other molecules of the same kind? Which molecule will undergo only London dispersion forces when interacting with other molecules of the same kind?

A. C_{4}H_{10}

B. NaC_{2}H_{3}O_{2}

C. CH_{2}C_{12}

D. C_{2}H_{5}OH HF

Learn more about the London dispersion forces here,

brainly.com/question/22388709

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Answer:

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