Answer: Option (E) is the correct answer.
Explanation:
When we move from top to bottom in a group then there occurs an increase in atomic size of the atoms due to increase in the number of electrons.
For example, in group 2A elements beryllium is the smallest in size whereas radium being at the bottom is the largest in size.
Also, atomic number of beryllium is 4 and atomic number of radium is 88.
Thus, we can conclude that out of the given options radium is the 2A element which has the largest atomic radius.
Answer:
H₂SO₄
Explanation:
Given data:
Number of moles of H₂SO₄ = 15 mol
Number of moles of Fe = 13 mol
Which reactant is limiting reactant = ?
Solution:
Chemical equation:
3H₂SO₄ + 2Fe → Fe₂(SO₄)₃ + 3H₂
now we will compare the moles reactant with product.
H₂SO₄ : Fe₂(SO₄)₃
3 : 1
15 : 1/3×15 = 5
H₂SO₄ : H₂
3 : 3
15 : 15
Fe : Fe₂(SO₄)₃
2 : 1
13 : 1/2×13 = 6.5
Fe : H₂
2 : 3
13 : 3/2×13 = 19.5
Number of moles of product formed by H₂SO₄ are less thus it will act as limiting reactant.
Answer:
Cathode: Ag
Anode: Br₂
Explanation:
In the cathode must occur a reduction, so it's more likely to a metal atom be in the cathode. For the metals given the reduction reactions and the potential of reduction are:
Ag⁺ + e⁻ ⇒ Ag⁰ E° = + 0.80 V
Fe⁺² + 2e⁻ ⇒ Fe⁰ E° = - 0.44 V
Al⁺³ + 3e⁻ ⇒ Al⁰ E° = -1.66 V
As the potential for Ag is the higher, the reduction will occur for it first, so in the cathode will produce Ag.
For the anode an oxidation must occurs, so the reactions for the nonmetals are:
F₂ + 2e⁻ ⇒ 2F⁻ E° = +2.87 V
Cl₂ + 2e⁻ ⇒ 2Cl⁻ E° = +1.36 V
Br₂ + 2e⁻ ⇒ 2Br⁻ E° = +1.07 V
For oxidation, the less the E°, the faster the reaction will occur, so Br₂ will be formed in the anode.
Answer:
For a liquid, it is 25°F
Explanation:
The standard state for a liquid is 25°C
