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vagabundo [1.1K]
3 years ago
7

Gold is isolated from rocks by reaction with aqueous cyanide, CN:4 Au(s)+8 NaCN(aq)+O 2 (g)+H 2 O(l) longrightarrow 4 Na[Au(CN)

2 ](aq)+4 NaOH(aq) . (a) Which atoms from which compounds are being oxidized, and which at oms from which compounds are being reduced?
Chemistry
1 answer:
qwelly [4]3 years ago
8 0

Au gets oxidised and Oxygen gets reduced.

<u>Explanation:</u>

4Au(s) + 8NaCN(aq) + O₂(g) + H₂O(l) → 4 Na[Au(CN)₂](aq) + 4 NaOH(aq)

Au is in 0 oxidation state is converted to an compound in that Au is in +2 oxidation state so it gets oxidised.

O₂, gets reduced from 0 to -1 oxidation state, so it also gets reduced.

Na and CN remains as such but with different atoms.

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The first-order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochlori
Dvinal [7]

Answer:

K(48.5°C) = 1.017 E-8 s-1

Explanation:

  • CH3Cl + H2O → CH3OH + HCl

at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1

at T2 = 48.5°C (321.5 K) ⇒ K2 = ?

Arrhenius eq:

  • K(T) = A e∧(-Ea/RT)
  • Ln K = Ln(A) - [(Ea/R)(1/T)]

∴ A: frecuency factor

∴ R = 8.314 E-3 KJ/K.mol

⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)

⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)

(1)/(2):

⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)

⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)

⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)

⇒ Ln (K1/K2) = - 3.422

⇒ K1/K2 = e∧(-3.422)

⇒ (3.32 E-10 s-1)/K2 = 0.0326

⇒ K2 = (3.32 E-10 s-1)/0.0326

⇒ K2 = 1.017 E-8 s-1

7 0
3 years ago
What is the mass of air in room
tigry1 [53]

Answer:

the mass of the air in the room is 4.96512 kg ( in 0°C)

3 0
3 years ago
PLZ HELP, Its The Nature of Oxidation and Reduction for chemistry
elixir [45]

Answer:

Oxidation is defined as the chemical process in which substance loses electron and hydrogen or gain oxygen while in the process of reduction, substance gains electron and hydrogen or loses oxygen.

So, from the given equation:

a. It is an oxidation reaction as Rb loses one elctron.

b. It is a reduction reaction as Te gains two electrons and become Te2-

c. It is a reduction reaction as H atom gains electrons.

d. It is an oxidation reaction as P loses 3 electrons.

8 0
3 years ago
6832 J of heat energy is applied to 5.9 mol of water. If the original temperature of the water was 18.60C, the final temperature
sineoko [7]

Answer: The final temperature of the water will be 34.0^0C

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

Q=m\times c\times \Delta T

Q = Heat absorbed=6832 Joules

m= mass of water = 5.9mol\times 18g/mol=106.2g

c = specific heat capacity = 4.184J/g^0C

Initial temperature of the water = T_i = 18.6^0C

Final temperature of the water = T_f  = ?

Putting in the values, we get:

6832=106.2\times 4.184\times (T_f-18.6)

T_f=34.0^0C

The final temperature of the water will be 34.0^0C

8 0
3 years ago
Explain how to use the periodic table to deduce the number of protons, neutrons and electrons of an atom of a specific element
Ksju [112]

Answer:

See explanation

Explanation:

The periodic table shows the atomic number and mass number of each element.

We know that the atomic number shows;

  1. The number of protons in the nucleus of the atom
  2. The number of electrons in the neutral atom of the element.

So we obtain the number of protons and electrons by looking at the atomic number shown in the periodic table.

We also know that;

Mass number = Number of protons + number of neutrons

Since number of protons = atomic number of the atom

Number of neutrons = Mass number - atomic number

Hence we obtain the number of protons by subtracting the atomic number from the mass number given in the periodic table.

5 0
3 years ago
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