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Julli [10]
1 year ago
13

Why do whales absorb CO2?

Chemistry
1 answer:
True [87]1 year ago
7 0

Answer:

Whales facilitate carbon absorption in two ways. On the one hand, their movements — especially when diving — tend to push nutrients from the bottom of the ocean to the surface, where they feed the phytoplankton and other marine flora that suck in carbon, as well as fish and other smaller animals.

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What is the process in which alkene monomers join end-to-end, or add on to one another, to form chains?
Thepotemich [5.8K]
ALKENES LIKE ETHYLENE JOIN END TO END TO FORM POLYETHYLENE BY ADDITION POLYMERIZATION TO FORM A LONG CHAIN SINCE THEY ARE UNSATURATED.
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8 0
3 years ago
Read 2 more answers
Be sure to answer all parts. the equilibrium constant (kp) for the reaction below is 4.40 at 2000. k. h2(g) + co2(g) ⇌ h2o(g) +
svetoff [14.1K]
When we have the balanced reaction equation is:

H2(g)  + CO2(g) ↔ H2O(g)  + CO (g)

a) first, to calculate ΔG° for the reaction: 

we will use this formula:

ΔG° = -RT㏑Kp

when R is R- rydberg constant = 8.314J/mol.K

and T is the temperature in Kelvin = 2000 K

and Kp = 4.4 

so, by substitution:

ΔG° = - 8.314 *2000 *㏑4.4

        = - 24624 J/mol = - 24.6 KJ/mol

b) to calculate ΔG so, we will use this formula:

ΔG = ΔG° + RT㏑Qp

So we need first, to get Qp from the reaction equation:

when Qp = P products / P reactants

                 =  PH2O*PCO / PH2 * PCO2

                 = (0.66 atm * 1.2 atm) / (0.25 * 0.78)

                 = 4.1 

so by substitution:

ΔG = -24624 + 8.314*2000*㏑4.1

      = -1162 J/mol = - 1.16 KJ/mol
7 0
4 years ago
Read 2 more answers
Carbon dioxide gas occupies a volume of 20.0 L at 308K and 2.30 atm. What volume would it occupy at 416 K and 5.40 atm?
Brrunno [24]

Answer : The final volume of gas is, 11.5 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 2.30 atm

P_2 = final pressure of gas = 5.40 atm

V_1 = initial volume of gas = 20.0 L

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 308 K

T_2 = final temperature of gas = 416 K

Now put all the given values in the above equation, we get:

\frac{2.30atm\times 20.0L}{308K}=\frac{5.40atm\times V_2}{416K}

V_2=11.5L

Therefore, the final volume of gas is, 11.5 L

7 0
3 years ago
Read 2 more answers
Please answer ASAP!!
Alenkasestr [34]

Answer:

C.0.28 V  

Explanation:

Using the standard cell potential we can find the standard cell potential for a voltaic cell as follows:

The most positive potential is the potential that will be more easily reduced. The other reaction will be the oxidized one. That means for the reactions:

Cu²⁺ + 2e⁻ → Cu E° = 0.52V

Ag⁺ + 1e⁻ → Ag E° = 0.80V

As the Cu will be oxidized:

Cu → Cu²⁺ + 2e⁻

The cell potential is:

E°Cell = E°cathode(reduced) - E°cathode(oxidized)

E°cell = 0.80V - (0.52V)

E°cell = 1.32V

Right answer is:

<h3>C.0.28 V </h3>

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4 0
3 years ago
The fire department needs information on friction losses occurring between a water main and an open fire hydrant. At maximum mai
lisov135 [29]

Explanation:

The given data is as follows.

      P_{1} = 85 psig,   P_{2} = P_{atm} = 15 psia

        Q = 1620 gpm,  d = 2.5 inch,     l = 8 ft = 2.4384 m

According to Darey-Weisbach equation,

                        h_{l} = \frac{4fl \nu^{2}}{2gD}  ......... (1)

Value of 'f' will be decided on the basis of Reynold number.

As, it is known that R_{l} = \frac{\rho \nu d}{\mu}

where,  \mu_{water} = 10^{-3} kg/ms

As it is known that 1 gpm = \frac{1}{3.67} m^{3}/hr

So,  1 m^{3}/hr = 3.67 gpm

Therefore,   Q = 1620 \times \frac{1}{3.67}

                        = 441.4168 m^{3}/hr

                         = 0.1226 m^{3}/s

In, 1 inch = 2.54 cm = 0.0254 m

Therefore, d = 2.5 \times 0.0254 = 0.0635 m

                V = \frac{Q}{\frac{\pi}{4}d^{2}}

                    = \frac{0.1226}{0.785 \times (0.0635)^{2}}

                    = 38.73 m/s

Hence, we will calculate Reynold number as follows.

             R_{l} = \frac{1000 \times 38.73 \times 0.0635}{10^{-3}}

                             = 2459355

As R_{l} > 2000 then, it means that flow is turbulent.

As, f = 0.079 R^{-0.25}_{l}

        = 0.001994

Putting all the values into equation (1) formula as follows.    

                          h_{l} = \frac{4fl \nu^{2}}{2gD}

                                     = \frac{4 \times 0.001994 \times 2.4384 \times (38.73)^{2}}{2 \times 9.81 \times 0.0635}

                                      = 1.04069 \times 10^{5} m

Thus, we can conclude that friction loss from the main to the discharge point is 1.04069 \times 10^{5} m.

4 0
3 years ago
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