Can someone help me please?

bartleby whihch of the foololwing is the correct cell notation for the reaction Hg2 Cd(s)-> Cd2 2Hg(l)

Misha Larkins [42]

Cd2+ + 2Hg Cd + Hg22+. Both Cd2+ + 2e Cd(s) -0.40 and Hg22+ + 2e 2Hg(l) 0.79

A chemical reaction known as an oxidation-reduction (redox) reaction includes the exchange of electrons between two substances.

Any chemical reaction in which the oxidation number of a molecule, atom, or ion changes by acquiring or losing an electron is referred to as an oxidation-reduction reaction. Decomposition Reaction is one of the several redox reactions.

This is the redox reaction's overall cell potential. Cd2+ + 2Hg Cd + Hg22+. Both Cd2+ + 2e Cd(s) -0.40 and Hg22+ + 2e 2Hg(l) 0.79

Reduction describes the increase in electrons. Oxidation and reduction always occur jointly because any loss of electrons by one substance must be followed by a gain of electrons by another.

Therefore, oxidation-reduction processes or simply redox reactions are other names for electron-transfer events.

Learn more about redox reactions here brainly.com/question/8727728.

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3 0

1 year ago

The units of density are kg/m2. If the density of a liquid is 760.0 kg/m' what is the specific volume? a) 1.316 x 10 m2/kg b) 1.

zlopas [31]

Answer:

The correct answer is: 1.316 . 10⁻³ m³/kg.

Explanation:

The density (ρ) of a substance is the ratio of its mass (m) to its volume (V). At constant temperature and pressure, its value is constant and it is an intrinsic property of materials. The units of density are kg/m³.

$\rho = \frac{m}{V}$

The specific volume (ν) of a substance is the ratio of its mass to its volume. We can see that it is the reciprocal of density and an intrinsic property of matter as well. Therefore, the units of specific volume are m³/kg.

$\nu = \frac{V}{m}=\frac{1}{\rho }$

Given we know the density of the liquid, we can use this relationship to find out its specific volume:

$\nu =\frac{1}{\rho }=\frac{1}{760.0kg/m^{3} } =1.316 .10^{-3} m^{3} /kg$

6 0

3 years ago

An automobile tire contains air at 320.×103 Pa at 20.0 ◦C. The stem valve is removed and the air is allowed to expand adiabatica

NISA [10]

Answer:

6.15.3 k

Explanation:

From the question we can see that

q = 0, Δu = w

Then,

$T_f = \frac{C_{V,m}+RP_{ext}P_i}{C_{V,m}+RP_{ext}P_f} T_i$

putting values wet

= $\frac{2.5\times 8.314+8.314\left(10^5\right)\left(3.20\times 10^5\right)}{2.5\times 8.314+\left(8.314\right)\left(10^5\right)\left(10^5\right)}\times \:293$