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2 years ago

How much negative charge is on each oxygen of the carbonate ion?

Chemistry

1 answer:

olga_2 [115]2 years ago

5 0

Answer:

Applying this same method to the carbonate ion, we have 3 resonance structures with bond orders of 2, 1, 1 when considering the bond between carbon and a single oxygen.

Explanation:

CO32- has three resonating structures. The way to calculate the bond order for most conpounds having resonating structures is:(No.of bonds/No.of resonating structures) Since, CO32- has 4 bonds, Thus answer will be 4/3 or 1.3333..

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What were the first galaxies made of

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3 years ago

The pH of an unknown strong acid is 2. What<br> is the hydronium ion concentration?

MissTica

Answer: Finding the [H3O+] and pH of Strong and Weak Acid Solutions The larger the Ka, the stronger the acid and the higher the H+ concentration at equilibrium. hydronium ion, H3O+, 1.0, 0.00, H2O, 1.0×10−14, 14.00.

Explanation:The hydrogen ion in aqueous solution is no more than a proton, a bare ... the interaction between H+ and H2O .

8 0

3 years ago

Materials expand when heated. Consider a metal rod of length L0 at temperature T0. If the temperature is changed by an amount ΔT

marysya [2.9K]

Answer:

(a) The length at temperature 180°C is 40.070 cm

(b) The length at temperature 90°C is 64.976 inches

Explanation:

(a) The given parameters are

The thermal expansion coefficient, α for steel = 1.24 × 10⁻⁵/°C

The initial length of the steel L₀ = 40 cm

The initial temperature, t₀ = 40°C

The length at temperature 180°C = L

Therefore, from the given relation, for change in length, ΔL, we have;

ΔL = α × L₀ × ΔT

The amount the temperature changed ΔT = 180°C - 40°C = 140°C

Therefore, the change in length, ΔL, is found as follows;

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 40 × 140°C = 0.07 cm

Therefore, L = L₀ + ΔL = 40 + 0.07 = 40.07 cm

The length at temperature 180°C = 40.07 cm

(b) Given that the length at T = 120°C is 65 in., we have;

The temperature at which the new length is sought = 90°C

The amount the temperature changed ΔT = 90°C - 120°C = -30°C

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 65 × -30°C = -0.024375 inches

The length, L at 90°C is therefore, L = L₀ + ΔL = 65 - 0.024375 = 64.976 in.

The length at temperature 90°C = 64.976 inches

Therefore;

L = 60 + α × 60 × (T - 150°C)

L = 60 + α × 60 × T - 9000 × α

L(T, α) = 60·α·T - 9000·α + 60

7 0

3 years ago

Calculate for the following electrochemical cell at 25°C, Pt H2(g) (1.0 atm) H (0.010 M || Ag (0.020 M) Ag if E (H) - +0.000 V a

viva [34]

Answer : The correct option is, (b) +0.799 V

Solution :

The values of standard reduction electrode potential of the cell are:

$E^0_{[H^{+}/H_2]}=+0.00V$

$E^0_{[Ag^{+}/Ag]}=+0.799V$

From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : $H_2\rightarrow 2H^{+}+2e^-$