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:
V₂ = 70.80 L
Explanation:
Given data:
Initial volume of He = 45 L
Initial temperature = 250 °C
Final temperature = 550 °C
Final volume = ?
Solution:
Initial temperature = 250 °C (250+273.15 K= 523.15 K)
Final temperature = 550 °C (550 + 273.15 k= 823.15 K)
According to Charles Law.,
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 45 L × 823.15 K / 523.15 K
V₂ = 37041.75 L.K / 523.15 K
V₂ = 70.80 L
Answer:
See the answer below
Explanation:
Beryllium forms an ion with 2 electrons in its outermost shell. Hence<u> the charge on the element is 2+</u>, that is, Be2+. In order for Be to form a stable bond with another element, <u>such an element must have a deficit of electrons in its outermost shell</u>. Beryllium will have to donate its 2 electrons to this element for its self-stability.
Answer:
a) = 0.704%
b) = 1.30%
c) = 2.60%
Explanation:
Given that:
=
For Part A; where Concentration of A = 0.270 M
Percentage Ionization(∝)
percentage% (∝) =
= 0.704%
For Part B; where Concentration of B = M
percentage% (∝) = 0.0130 × 100%
= 1.30%
For Part C; where Concentration of C=
percentage% (∝) = 0.02608 × 100%
= 2.60%