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4 years ago

Which solution has a molality of 0.25m nacl?

2 answers:

5 0

Molality can be expressed by moles of solute over kilograms of solvent. The question asks the molality of 0.25m NaCl. 0.25m NaCl is equal to 0.25 moles of NaCl over 1 kg of water.

vodomira [7]4 years ago

4 0

0.25 moles of NaCl over 1 kg of water

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g Calculate the time (in min.) required to collect 0.0760 L of oxygen gas at 298 K and 1.00 atm if 2.60 A of current flows throu

lozanna [386]

Answer:

7.67 mins.

Explanation:

Data obtained from the question include the following:

Volume (V) = 0.0760 L

Temperature (T) = 298 K

Pressure (P) = 1 atm

Current (I) = 2.60 A

Time (t) =?

Next, we shall determine the number of mole (n) of O2 contained in 0.0760 L.

This can be obtained by using the ideal gas equation as follow:

Note:

Gas constant (R) = 0.0821 atm.L/Kmol

PV = nRT

1 x 0.0760 = n x 0.0821 x 298

Divide both side by 0.0821 x 298

n = 0.0760 / (0.0821 x 298)

n = 0.0031 mole

Next, we shall determine the quantity of electricity needed to liberate 0.0031 mole of O2.

This is illustrated below:

2O²¯ + 4e —> O2

Recall:

1 faraday = 1e = 96500 C

4e = 4 x 96500 C

4e = 386000 C

From the balanced equation above,

386000 C of electricity liberated 1 mole of O2.

Therefore, X C of electricity will liberate 0.0031 mole of O2 i.e

X C = 386000 X 0.0031

X C = 1196.6 C

Therefore, 1196.6 C of electricity is needed to liberate 0.0031 mole of O2

Next, we shall determine the time taken for the process. This can be obtained as follow:

Current (I) = 2.60 A

Quantity of electricity (Q) = 1196.6 C

Time (t) =?

Q = It

1196.6 = 2.6 x t

Divide both side by 2.6

t = 1196.6/2.6

t = 460.23 secs.

Finally, we shall convert 460.23 secs to minute. This can be achieved by doing the following:

60 secs = 1 min

Therefore,

460.23 secs = 460.23/60 = 7.67 mins

Therefore, the process took 7.67 mins.

7 0

4 years ago

For parts a & b below, derive only the initial value problem set up.

Otrada [13]

Answer:

Explanation:

From the given information:

Let y(t) be the amount of sugar in tank A at any time.

Then, the rate of change of sugar in the tank is given by:

$\dfrac{dy}{dt}= (rate \ in ) - ( rate \ out )$

The rate of the sugar coming into the tank is 0

$\text{rate of sugar going out }= \dfrac{y(t) \ pound}{100}\times \dfrac{1 }{min} \\ \\ = \dfrac{y}{100} \ pound/min$

$So; \\ \\ \\ \dfrac{dy}{dt} = 0 - \dfrac{y}{100} \\ \\ \implies \dfrac{dy}{y }= -\dfrac{dt}{100 } \\ \\ \implies dx|y| = -\dfrac{t}{100}+C \\ \\ \implies e*{In|y|}= e^{-\dfrac{t}{100}+C} \\ \\ \implies y = Ce^{-\dfrac{t}{100}}$

Initial amount of sugar = 25 Pounds

Now; y(0) = 25

25 = Ce⁰

C = 25

So; y(t) = 25 $e^{-\dfrac{t}{100}$

Thus, the amount of sugar at any time t = $\mathbf{25 e^{^{-\dfrac{t}{100}}}}$

B) For tank B :

$\dfrac{dy}{dt}= 1 - \dfrac{y}{50 } \\ \\ \dfrac{dy}{dt}= \dfrac{50-y}{50} \\ \\ \dfrac{dy}{50-y }= \dfrac{dt}{50}$

8 0

3 years ago

plz answer 7 & 8 because I dont get the answer... oh and you need to use the picture for the questions 7 & 8 thanks

djyliett [7]

7. An exothermic reaction
8. The bonds are forming

4 0

4 years ago

To function on their own, unicellular organisms must do what?

UNO [17]

Well you see unicellular organisms are very unique in the fact that they live all on their own and move around with a flagellum or little motor, but to the point unicellular organisms carry out essentially every function in life and they must in order to simply survive. So it's most likely D.

5 0

3 years ago

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Which of the following is a common property or characteristic of an acid?

Rina8888 [55]

The characteristic that is common to acids are that they can react with active metals to produce hydrogen gas.

6 0

3 years ago

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