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ELEN [110]
3 years ago
14

What is the pH of a 0.025 M [OH] solution?

Chemistry
1 answer:
Debora [2.8K]3 years ago
8 0
The answer is 12.4.I think its correct answer.

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What is the frequency of the sinusoidal graph?
Arisa [49]

Answer:

For finding frequency, we need to first find the period of the graph.

The period of a sinusoidal graph is the time interval in which it repeats its pattern.

In the graph, we can see, after 1/period time, it repeats its pattern.

Hence the period of the graph is  \pi /2.

Now we need to find its frequency 1/\pi /2

The formula for frequency is 2/\pi

This is the answer

i hope you pass the assignment

try your best!

8 0
3 years ago
You add 100.0 g of water at 52.0 °C to 100.0 g of ice at 0.00 °C. Some of the ice melts and cools the water to 0.00 °C. When the
lara31 [8.8K]

Answer:

m_{ice} = 65.336\,g

Explanation:

Accoding to the First Law of Thermodynamics, the heat released by the water melts a portion of ice. That is to say:

Q_{water} = Q_{ice}

(100\,g)\cdot \left(4.184\,\frac{J}{kg\cdot ^{\textdegree}C}\right)\cdot (52\,^{\textdegree}C - 0\,^{\textdegree}C) = m_{ice}\cdot \left(333\,\frac{J}{g} \right)

The amount of ice that is melt is:

m_{ice} = 65.336\,g

5 0
4 years ago
How many liters of H2O gas are produced when
anyanavicka [17]
1 mole C3H8 produces 4 moles H2O. So, first we convert 32 grams of propane to moles and then find moles of H2O. Then convert moles of H2O to grams of H2O
Moles of H2O produced = 32 g C3H8 x 1 mole/44 g x 4 moles H2O/mole C3H8 = 2.909 moles H2O
Grams H2O produced = 2.909 moles H2O x 18 g/mole = 52.36 g = 52 g H2O
8 0
2 years ago
A sample of hydrogen gas was collected over water. If the total pressure
Brut [27]
It would 47.7 because you would have to both minus the number together.
7 0
3 years ago
5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buf
mr_godi [17]

Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and pK_{a} is as follows.

               pH = pK_{a} + log \frac{base}{acid}

where,     pH = 7.4 and pK_{a} = 7.21

As here, we can use the pK_{a} nearest to the desired pH.

So,      7.4 = 7.21 + log \frac{base}{acid}

             0.19 = log \frac{base}{acid}

            \frac{base}{acid} = 1.55

1 mM phosphate buffer means [HPO_{4}] + [H_{2}PO_{4}] = 1 mM

Therefore, the two equations will be as follows.

           \frac{HPO_{4}}{H_{2}PO_{4}} = 1.55 ............. (1)

  [HPO_{4}] + [H_{2}PO_{4}] = 1 mM ........... (2)        

Now, putting the value of [HPO_{4}] from equation (1) into equation (2) as follows.

             1.55[H_{2}PO_{4}] + [tex][H_{2}PO_{4}] = 1 mM

                        2.55 [H_{2}PO_{4}] = 1 mM

                             [H_{2}PO_{4}] = 0.392 mM

Putting the value of [H_{2}PO_{4}] in equation (1) we get the following.

                     0.392 mM + [HPO_{4}] = 1 mM

                          [HPO_{4}] = (1 - 0.392) mM

                              [HPO_{4}] = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.

7 0
3 years ago
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