Cl2(s); oxidation number 1 is the incorrect choices in oxidation number.
Explanation:
In the elemental form oxidation state is zero. Here chlorine is present in elemental form so oxidation state is zero.
Oxidation number depends on the number of electrons gained or lost by an atom of the element say in compound formation.
If electron is gained oxidation number becomes negative.
If electron is lost then oxidation number is positive.
If the octet rule is fulfilled that valence shell is filled them atomic number gets zero. Since Cl2 is in neutral state the oxidation number is 0.
Oxidation number in general can be made out by checking the valency of the element as oxidation number is also equal to the valency.
NaOH but you didnt give options
Answer:
The answer to your question is given below
Explanation:
1. C. 2NaCl + I2 —> 2NaI + Cl2 => C. Single displacement.
From the above equation, we can see that I2 replaces Cl in NaCl to produce NaI. This is simply called a single displacement reaction.
2. E. 2C4H10 + 13O2 —> 8CO2 + 10H2O => E. Combustion.
The above equation shows the burning of Hydrocarbon in the presence of O2. This is simply called Combustion as CO2 and H2O is produced.
3. D 2H2O —> 2H2 + O2 => D. Decomposition.
From the above equation, we can see that a single compound H2O produces two elements H2 and O2. This is simply called a decomposition reaction.
4. A. ZnS + 2HCl —> ZnCl2 + H2S => A. Double Decomposition.
From the above equation, we can see that Cl replaces S in ZnS to produce ZnCl2 and S replaces Cl in HCl to produce H2S. This is simply called double displacement reaction.
5. B. H2 + Br2 —> 2HBr => B. Synthesis.
From the above equation, we can see that two element H2 and Br2 combine to produce a single compound HBr. This is simply called a synthesis reaction.
Answer:
-21 kJ·mol⁻¹
Explanation:
Data:
H₃O⁺ + OH⁻ ⟶ 2H₂O
V/mL: 50 50
c/mol·dm⁻³: 1.0 1.0
ΔT = 4.5 °C
C = 4.184 J·°C⁻¹g⁻¹
C_cal = 50 J·°C⁻¹
Calculations:
(a) Moles of acid
So, we have 0.050 mol of reaction
(b) Volume of solution
V = 50 dm³ + 50 dm³ = 100 dm³
(c) Mass of solution
(d) Calorimetry
There are three energy flows in this reaction.
q₁ = heat from reaction
q₂ = heat to warm the water
q₃ = heat to warm the calorimeter
q₁ + q₂ + q₃ = 0
nΔH + mCΔT + C_calΔT = 0
0.050ΔH + 100×4.184×4.5 + 50×4.5 = 0
0.050ΔH + 1883 + 225 = 0
0.050ΔH + 2108 = 0
0.050ΔH = -2108
ΔH = -2108/0.0500
= -42 000 J/mol
= -42 kJ/mol
This is the heat of reaction for the formation of 2 mol of water
The heat of reaction for the formation of mol of water is -21 kJ·mol⁻¹.
The soda an air is most likely bubbling up and is going to explode.
