water plus oxygen equals rust so keep water away from the iron to prevent rusting or dry the iron off then apply alchohol to cleanse it
Answer:
Empirical formula is CH₄
Molecular formula = C₂H₈
Explanation:
Mass of carbon = 37.5 g
Mass of hydrogen = 12.5 g
Molecular weight = 32 g/mol
Molecular formula = ?
Empirical formula = ?
Solution:
Number of gram atoms of C = 37.5 g /12g/mol = 3.125
Number of gram atoms of H = 12.5 g / 1.008 g/mol= 12.4
Atomic ratio:
C : H
3.125/3.125 : 12.4 /3.125
1 : 4
C : H : = 1 : 4
Empirical formula is CH₄
Molecular formula:
Molecular formula = n (empirical formula)
n = molar mass of compound / empirical formula mass
n = 32 / 16
n = 2
Molecular formula = n (empirical formula)
Molecular formula = 2 ( CH₄)
Molecular formula = C₂H₈
Answer:
reactive nonmetals since they have a full valence shell (that's why they're stable).
Answer:
AWWWWWWWWWWWWWWWW SOOO NICEEE
The mass of ammonium chloride that must be added is : ( A ) 4.7 g
<u>Given data :</u>
Volume of water ( V ) = 250 mL = 0.25 L
pH of solution = 4.85
Kb = 1.8 * 10⁻⁵
Kw = 10⁻¹⁴
Given that the dissolution of NH₄Cl gives NH₄⁺⁺ and Cl⁻ ions the equation is written as :
NH₄CI + H₂O ⇄ NH₃ + H₃O⁺
where conc of H₃O⁺
[ H₃O⁺ ] =
and Ka = Kw / Kb
∴ Ka = 5.56 * 10⁻¹⁰
Next step : Determine the concentration of H₃O⁺ in the solution
pH = - log [ H₃O⁺ ] = 4.85
∴ [ H₃O⁺ ] in the solution = 1.14125 * 10⁻⁵
Next step : Determine the concentration of NH₄CI in the solution
C = [ H₃O⁺ ]² / Ka
= ( 1.14125 * 10⁻⁵ )² / 5.56 * 10⁻¹⁰
= 0.359 mol / L
Determine the number of moles of NH₄CI in the solution
n = C . V
= 0.359 mol / L * 0.25 L = 0.08979 mole
Final step : determine the mass of ammonium chloride that must be added to 250 mL
mass = n * molar mass
= 0.08979 * 53.5 g/mol
= 4.80 g ≈ 4.7 grams
Therefore we can conclude that the mass of ammonium chloride that must be added is 4.7 g
Learn more about ammonium chloride : brainly.com/question/13050932