The required moles of CaO are needed to react with an excess of water to form 370 grams of calcium hydroxide is 280g.
Relation between the mass and moles of any substance will be represented as:
n = W/M, where
Given chemical reaction is:
CaO + H₂O → Ca(OH)₂
Moles of Ca(OH)₂ = 370g / 74g/mol = 5mol
From the stoichiometry of the reaction, 5 moles of CaO is required to produce 5 moles of Ca(OH)₂.
Mass of CaO = (5mol)(56g/mol) = 280g.
Hence required mass of CaO is 280g.
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B. metalloid.
C. nonmetal.
D. transition metal.
Answer:
C) Neutralizing
Explanation:
A neutralizing shampoo is specifically formulated to restore the pH balance of the hair by neutralizing any alkali or unwanted residues that may have accumulated. It is commonly used after chemical treatments such as perms or relaxers, which can alter the hair's pH level.
Chemical processes like perms and relaxers often involve the use of alkaline substances to break down and reshape the hair structure. After the process is complete, a neutralizing shampoo is used to stop the chemical reaction and bring the hair's pH back to its normal, balanced state. This helps to ensure the hair's health, smoothness, and manageability.
According to the mole concept, there are 1.8074 moles in 77.1 g of chlorine.
Mole is defined as the unit of amount of substance . It is the quantity measure of amount of substance of how many elementary particles are present in a given substance.
It is defined as exactly 6.022×10²³ elementary entities. The elementary entity can be a molecule, atom ion depending on the type of substance. Amount of elementary entities in a mole is called as Avogadro's number.
It is widely used in chemistry as a suitable way for expressing amounts of reactants and products.For the practical purposes, mass of one mole of compound in grams is approximately equal to mass of one molecule of compound measured in Daltons. Number of moles is calculated as, mass/molar mass
Substituting values in formula gives, number of moles=77.1/70.90=1.8074
Thus, there are 1.8074 moles in 77.1 g of Cl₂.
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Answer:
Explanation:
We must look up the standard reduction potentials for the half-reactions.
ℰ°
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⇌ 2Cr³⁺ + 7H₂O 1.36
Cl₂ + 2e⁻ ⇌ 2Cl⁻ 1.35827
2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O 1.195
Fe³⁺ + e⁻ ⇌ Fe²⁺ 0.771
(a) Cr³⁺/Cl₂
We reverse the sign of ℰ° for the oxidation half-reaction. Then we add the two half-reactions and their ℰ° values.
ℰ°/V
2Cr³⁺ + 7H₂O ⇌ Cr₂O₇²⁻ + 14H⁺ + 6e⁻ -1.36
Cl₂ + 2e⁻ ⇌ 2Cl⁻ 1.358 27
2Cr³⁺ + 3Cl₂ + 7H₂O ⇌ Cr₂O₇²⁻ + 6Cl⁻ + 14H⁺ 0.00
(b) Fe²⁺/IO₃⁻
ℰ°/V
Fe²⁺ ⇌ Fe³⁺ + e⁻ -0.771
2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O 1.195
10Fe²⁺ + 2IO₃⁻ + 12H⁺ ⇌ 10Fe³⁺ + I₂ + 6H₂O 0.424
The ℰ° values for the cells are
Answer:
Answer:
\boxed{\text{(a) 0.00 V; (b) 0.424 V}}
Explanation:
We must look up the standard reduction potentials for the half-reactions.
ℰ°
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⇌ 2Cr³⁺ + 7H₂O 1.36
Cl₂ + 2e⁻ ⇌ 2Cl⁻ 1.35827
2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O 1.195
Fe³⁺ + e⁻ ⇌ Fe²⁺ 0.771
(a) Cr³⁺/Cl₂
We reverse the sign of ℰ° for the oxidation half-reaction. Then we add the two half-reactions and their ℰ° values.
ℰ°/V
2Cr³⁺ + 7H₂O ⇌ Cr₂O₇²⁻ + 14H⁺ + 6e⁻ -1.36
Cl₂ + 2e⁻ ⇌ 2Cl⁻ 1.358 27
2Cr³⁺ + 3Cl₂ + 7H₂O ⇌ Cr₂O₇²⁻ + 6Cl⁻ + 14H⁺ 0.00
(b) Fe²⁺/IO₃⁻
ℰ°/V
Fe²⁺ ⇌ Fe³⁺ + e⁻ -0.771
2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O 1.195
10Fe²⁺ + 2IO₃⁻ + 12H⁺ ⇌ 10Fe³⁺ + I₂ + 6H₂O 0.424
The ℰ° values for the cells are \boxed{\textbf{(a) 0.00 V; (b) 0.424 V}}
Explanation:
its right trust