Explanation:
0.00208 is the correct standard form
The British Petroleum (bp's) effort to close the blowout preventer and install a containment dome following an explosion on the deepwater horizon drilling rig is an example of <u>regulatory policy </u>change.
The deepwater horizon drilling rig is a semi-submersible, transportable, floating, flexibly oriented drilling rig that could work in sea depths of up to 10,000 feet which is approximately 3,000 meters.
During the explosion that occurred in April 2010, British Petroleum made several efforts to contain the damages made and to prevent further outbreaks of disasters.
Part of the changes was shifting from a blowout preventer that has a specialized valve to seal, manage, and monitor oil and gas wells in order to prevent blowouts to a containment dome (a crucial component of a system meant to control an oil well's underwater blowout).
Therefore, we can conclude that the British Petroleum (bp's) effort to close the blowout preventer and install a containment dome following an explosion on the deepwater horizon drilling rig is an example of <u>regulatory policy </u>change.
Learn more about deepwater horizon drilling rig here:
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Answer:
the region of negative charge surrounding an atomic nucleus that is associated with an atomic orbital. The region is defined mathematically, describing a region with a high probability of containing electrons.
Explanation:
<span>9.40x10^19 molecules.
The balanced equation for ammonia is:
N2 + 3H2 ==> 2NH3
So for every 3 moles of hydrogen gas, 2 moles of ammonia is produced. So let's calculate the molar mass of hydrogen and ammonia, starting with the respective atomic weights:
Atomic weight nitrogen = 14.0067
Atomic weight hydrogen = 1.00794
Molar mass H2 = 2 * 1.00794 = 2.01588 g/mol
Molar mass NH3 = 14.0067 + 3 * 1.00794 = 17.03052 g/mol
Moles H2 = 4.72 x 10^-4 g / 2.01588 g/mol = 2.34140921086573x10^-4 mol
Moles NH3 = 2.34140921086573x10^-4 mol * (2/3) = 1.56094x10^-4 mol
Now to convert from moles to molecules, just multiply by Avogadro's number:
1.56094x10^-4 * 6.0221409x10^23 = 9.400197448261x10^19
Rounding to 3 significant figures gives 9.40x10^19 molecules.</span>
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
