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3 years ago

What was the company responsible for the massive oil spill in the Gulf of Mexico in April 2010?

Chemistry

2 answers:

sleet_krkn [62]3 years ago

5 0

Answer:

Transocean

Explanation:

pickupchik [31]3 years ago

3 0

Answer:Transocean

Explanation:

The Deepwater Horizon rig, owned and operated by offshore-oil-drilling company Transocean and leased by oil company BP, was situated in the Macondo oil prospect in the Mississippi Canyon, a valley in the continental shelf.

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Please solve quickly!!!!

ohaa [14]

6 ( tell me if you got it right)

3 0

3 years ago

in a certain solvent is first order with respect to (CH3)3CBr and zero order with respect to OH2. In several experiments the rat

Inga [223]

Answer:

Ea= -175.45J

A= 3.5×10^14

k=3.64 ×10^14 s^2.

Explanation:

From

ln k= -(Ea/R) (1/T) + ln A

This is similar to the equation of a straight line:

y= mx + c

Where m= -(Ea/R)

c= ln A

y= ln k

Therefore

21.10 3 104= -(Ea/8.314)

Ea=-( 21.10 3 104×8.314)

Ea= -175.45J

b) ln A= 33.5

A= e^33.5

A= 3.5×10^14

k= Ae^-Ea/RT

k= 3.5×10^14 × e^ -(-175.45/8.314×531)

k = 3.64 ×10^14 s^2.

8 0

3 years ago

Which subshells (s, p, d, f, or g) can have electrons with the indicated magnetic quantum number (ml)?

amm1812

Answer:

=3 means is 3 or greater so that would be f and g subshells

=0 means is 0 or greater so that would be s, p, d, f and g subshells

=1 means is 1 or greater so that would be p, d, f, and g subshells

=4 means is 4 or greater so that would be g only

4 0

2 years ago

Can someone please help it is due tonight!

kompoz [17]

Reactants left products right

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3 years ago

Read 2 more answers

Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow

andreyandreev [35.5K]

Explanation:

(a) The given data is as follows.

Pressure on top ( $P_{o}$ ) = 140 bar = $1.4 \times 10^{7} Pa$ (as 1 bar = $10^{5}$ )

Temperature = $15^{o}C$ = (15 + 273) K = 288 K

Density of gas = $\frac{PM}{ZRT}$

$\frac{dP}{dZ} = \rho \times g$

$\frac{dP}{dZ} = \int \frac{PM}{ZRT}$

$\int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ$

$ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}$

= 0.4548

$P_{1} = P_{o} \times e^{0.4548}$

= $1.4 \times 10^{7} Pa \times 1.5797$

= $2.206 \times 10^{7} Pa$

Hence, pressure at the natural gas-oil interface is $2.206 \times 10^{7} Pa$ .

(b) At the bottom of the tank,

$P_{2} = P_{1} + \rho \times g \times h$

= 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

= $309.8 \times 10^{5} Pa$

= 309.8 bar

Hence, at the bottom of the well at $15^{o}C$ pressure is 309.8 bar.

6 0

3 years ago