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Nikolay [14]
3 years ago
5

What problems might there be if oxygen was very soluble in water?​

Chemistry
1 answer:
e-lub [12.9K]3 years ago
8 0
There is a lot of water on earth. If oxygen was very soluble in water, the oceans would absorb and hold the oxygen of the atmosphere and there would be none left for plants and animals.

Hope this help you!!
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What cells do you think is in metaphase? why?
Tcecarenko [31]
Etaphase is a stage during the process of cell division (mitosis or meiosis). Usually, individual chromosomes cannot be observed in the cell nucleus. However, during metaphase of mitosis or meiosis the chromosomes condense and become distinguishable as they align in the center of the dividing cell.
3 0
2 years ago
A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path
Igoryamba

Explanation:

  • For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

              R = 8.314 J/(K mol),          T = 298 K ,

          V_{2} = 8.34 L,    V_{1} = 2.67 L

Now, putting the given values into the above formula as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

   = -2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})

     = -2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494

    = -2818.68 J

Hence, work for path A is -2818.68 J.

  • For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure P_{external} = 1.00 atm.

So,              W = -P_{external} \times \Delta V

Now, putting the given values into the above formula as follows.

               W = -P_{external} \times \Delta V

                   = -1 atm \times (8.34 L - 2.67 L)  

                    = -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

       W = -\frac{101.33 J}{1 atm L} \times 5.67 atm L

            = -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

                        W = -574.54 J

Hence, work for path B is -574.54 J.

3 0
4 years ago
What are the variables of Boyle's law?
Llana [10]
Boyles law states P1V1=P2V2
P being pressure and V being volume
So volume and pressure
6 0
3 years ago
Read 2 more answers
SC. 8.P.8.9
ELEN [110]

Answer:

C

Explanation:

8 0
3 years ago
If 11g of a gas occupies 5.6dm'3at s.t.p., calculate it's vapour density (1.0mol of a gas occupies 22.4dm'3at s.t.p.).​
11111nata11111 [884]

Answer:

{ \boxed{ \bf{vapour \: density = 2 \times molecular \: mass}}} \\{ \tt{ PV= (\frac{m}{ m_{r}}) RT}} \\ { \tt{3 \times 5.6 =  \frac{11}{m _{r}}  \times 0.0831 \times 273}} \\ { \tt{m _{r} = 14.85 \: g}} \\  \\ { \bf{vapour \: density = 2 \times m _{r}}} \\  = 2 \times 14.85 \\  = 29.7 \: { \tt{g {dm}^{ - 3} }}

7 0
3 years ago
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