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sweet-ann [11.9K]
3 years ago
10

When ammonium chloride is added to water and stirred, it dissolves spontaneously and the resulting solution feels cold. Without

doing any calculations, deduce the signs of ΔG, ΔH, and ΔS for this process, and justify your choices
Chemistry
1 answer:
d1i1m1o1n [39]3 years ago
7 0
When ammonium chloride NH4Cl is added to water and stirred, it dissolves spontaneously (this is the basis for ΔG) for and the resulting solution feels cold (endothermic, the basis for ΔH). Without doing any calculations, we can easily deduce the signs of ΔG, ΔH, and ΔS for this process based on the observations.

ΔG < 0 (it is spontaneous)
ΔH < 0 (because the process is endothermic - it absorbs energy)
ΔS > 0 (entropy increases because of the dissolution of NH4Cl in water
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What method of heat transfer takes place when particles of matter move from one place to another in a liquid
Alecsey [184]

Answer:

CONVECTION

Explanation:

CONVECTION: In liquids and gases, convection is usually the most efficient way to transfer heat. Convection occurs when warmer areas of a liquid or gas rise to cooler areas in the liquid or gas.

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3 years ago
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A 55.0 mL aliquot of a 1.50 M solution is diluted to a total volume of 278 mL. A 139 mL portion of that solution is diluted by a
balu736 [363]

Answer:

0.14 M

Explanation:

To determinate the concentration of a new solution, we can use the equation below:

C1xV1 = C2xV2

Where C is the concentration, and V the volume, 1 represents the initial solution, and 2 the final one. So, first, the initial concentration is 1.50 M, the initial volume is 55.0 mL and the final volume is 278 mL

1.50x55.0 = C2x278

C2 = 0.30 M

The portion of 139 mL will be the same concentration because it wasn't diluted or evaporated. The final volume will be the volume of the initial solution plus the volume of water added, V2 = 139 + 155 = 294 mL

Then,

0.30x139 = C2x294

C2 = 0.14 M

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3 years ago
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Write a balanced nuclear equation for the formation of 28 Si 14 from beta-minus emission.
Fantom [35]

Answer:

Phosphorus-28 undergoes beta-minus decay to produce

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\rm ^{28}_{13} Al \to ^{28}_{14} Si + ^{\phantom{-}0}_{-1}e + \bar{\mathnormal{v}}_{\rm e}

Explanation:

In simple words, when a nucleus undergoes beta-minus decay, a neutron is converted to a proton. An electron and an electron antineutrino will be released.

\rm ^{1}_{0}n^{0} \to ^{1}_{1}p^{+} + ^{\phantom{-}0}_{-1}e^{-} + \bar{\mathnormal{v}}_{\rm e}.

One way to tell whether a neutron is converted to a proton, but not vice versa, is to check the sum charges on the two sides of this equation.

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  • Right-hand side: 1 + (-1) = 0. Each proton carries a charge of +1. Each electron (beta-minus particle) carries a charge of -1. Antineutrinos are neutral.

The charges on the two sides of this equation is the same. Hence this nuclear equation is possible (but not necessarily correct; however, if the proton and the neutron are in the wrong place the charge won't even be the same.)

Since the mass number of a proton and a neutron are both 1, the overall mass number of the atom will stay the same.

The atomic number is the number of protons in each atom. That number determines the symbol and the chemical properties of the atom. When one neutron in an atom is converted to a proton, the atomic number of the atom will increase by 1.

The atomic number of the daughter nucleus, silicon, is 14. It takes a parent nucleus with atomic number 14 - 1 = 13 to produce a silicon atom. Refer to a modern periodic table. Atomic number 13 corresponds to the element aluminum.

Also, the mass number of the daughter nucleus is 28. Since the mass number would stay the same in a beta decay, the mass number of the parent nucleus would also be 28. In other words, it takes an aluminum-28 atom to undergo beta-decay to produce a silicon-28 atom.

Complete the other details (electron and electron antineutrino) to obtain the equation

\rm ^{28}_{13} Al \to ^{28}_{14} Si + ^{\phantom{-}0}_{-1}e + \bar{\mathnormal{v}}_{\rm e}.

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2Na⁺(l) + 2e⁻ → 2Na(l).

Reaction of oxidation at anode(+): 2I⁻(l) → I₂(l) + 2e⁻.

The anode is positive and the cathode is negative.


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Answer:

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