<span>crunching and folds the rock at the boundary, lifts it up and leads to the formation of mountains </span>
If you are given the number of
moles of sodium bicarbonate, use the balanced chemical equation. The molar mass
of the acetic acid is 60 grams per mole. Using stochiometric balance.
Number of moles of acetic acid =
3 moles NaHCO3 (1 mol CH3COOH/1 mol NaHCO3) = 3 moles of acetic acid
Grams of acetic acid = 3 moles
of CH3COOH (60 g CH3COOH/1 mol CH3COOH) = <span>180 grams of acetic acid</span>
Answer: - 986.6 kj/mol
Explanation:
1) Equation given:
CaO(s) + H₂O (l) → Ca (OH)₂ (s) δh⁰ = −65.2 kj/mol
2) Standard enthalpies of formation given:
CaO, δhf⁰ = −635.6 kj/mol
H₂O, δhf⁰ = −285.8 kj/mol
3) Calculate the standard enthalpy of formation of Ca(OH)₂.
δh⁰ = ∑δh⁰f of products - ∑ δh⁰f of reactants
Using the mole coefficients of the balanced chemical equation:
δh⁰ = δh⁰f Ca(OH)₂ - δh⁰f CaO - δh⁰f H₂O
⇒ δh⁰f Ca(OH)₂ = δh⁰ + δh⁰f CaO + δh⁰f H₂O
⇒ δh⁰f Ca(OH)₂ = - 65.2 kj/mol − 635.6 kj/mol) − 285.8 kj/mol) = - 986.6 kj/mol.
Answer:
d =~ 5.8μm
d =~ 0.13 μm
Explanation:
when the doping concentrations are 5 × 10^15 cm^-3
d = v^-1/3 ; where d represent the distance between the atoms , and v represent the volume
d =1/ ∛v
d = 1/ ∛5 × 10^15
d = 1/ 170997.5
d = 5.85 × 10 ^ -6
d =~ 5.8μm
when the doping concentrations are 5 × 10^20 cm^-3
d = v^-1/3 ; where d represent the distance between the atoms , and v represent the volume
d =1/ ∛v
d = 1/ ∛5 × 10^20
using the principle of surds and standard forms, we have
d = 1/ ∛0.5 × 10^21
d = 1/7937005.26
d = 1.26 × 10 ^ -7
d = 0.126 × 10 ^ -6
d =~ 0.13 μm
