Answer:
0.46 V
Explanation:
The emf for the cell is given by:
Eº cell = Eº oxidation + Eº reduction
From the given balanced chemical equation, we can deduce that Fe²⁺ has been oxidized to Fe³⁺, and O reduced from 0 to negative 2, according to the half cell reactions:
4Fe²⁺ ⇒ Fe³⁺ + 4e⁻ oxidation
O₂ + 4H⁺ + 4 e⁻ ⇒ 2 H₂O reduction
From reference tables for the standard reduction potential, we get
Eº red Fe³⁺ / Fe²⁺ Eºred = 0.77 V
Eº red O₂ / H₂O Eºred = 1.23 V
Now all we need to do is change the sign of Eº reduction for the species being oxidized ( Fe²⁺ ) and add it to Eº reduction O₂:
Eº cell = Eº oxidation + Eº reduction = - (0.77 V ) + 1.23 V = 0.46 V
The answer is 267.93 g
Molar mass of CaBr2 is the sum of atomic masses of Ca and Br:
Mr(CaBr2) = Ar(Ca) + 2Ar(Br)
Ar(Ca) = 40 g/mol
Ar(Br) = 79.9 g/mol
Mr(CaBr2) = 40 + 2 * 79.9 = 199.8 g/mol
The percentage of Br in CaBr2 is:
2Ar(Br) / Mr(CaBr2) * 100 = 2 * 79.9 / 199.8 * 100 = 79.98%
Now make a proportion:
x g in 79.98%
335 g in 100%
x : 79.98% = 335 g : 100%
x = 79.98% * 335 g : 100%
x = 267.93 g
Answer:
if you are asking k then the round one is condenser
if not then its a stage clip
The total number of ions in 38.1 g of SrF₂ is 5.479 x 10²³.
<h3>What are ions?</h3>
Ions are the elements with a charge on them. It happens when they share electrons with other atoms to form a compound.
We have to calculate the total number of ions in 38.1 g of .
The molar mass of SrF₂ = 125.62 g/mol
The number of moles = 38.1 g of 1.0 mol / 125.62 = 0.30329 moles
Given that, total moles of SrF₂ ions in = 1.0 mol of + 2.0 moles of = 3.0 moles
Total moles of ions in 0.30329 moles of
= (0.30329 moles of SrF₂) x 3.0 / 1.0 = 0.90988 mol ions
We know that,
1.0 mole of ions = 6.023 x 10²³ ions
Thus, the number of total ions = ( 0.90988 mol ions) x 6.023 x 10²³ / 1.0 mol = 5.479 x 10²³ ions
Thus, the number of ions is in 38.1 g of 5.479 x 10²³ ions
To learn more about ions, refer to the link:
brainly.com/question/14295820
#SPJ4
Answer:
See explanation
Explanation:
The central atom in the perbromate ion is bromine. The chemical symbol of bromine is Br. There are no lone pairs around the central bromine atom. The ion is tetrahedral in shape hence we expect a bond angle of 109°. 27 which is the ideal tetrahedral bond angle. The actual bond angle of the prebromate ion is 109.5°. The perbromate ion is BrO4^-
The observed bond angle is very close to the ideal value because of the absence of lone pairs of electrons from the central atom in the ion.
