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BartSMP [9]
3 years ago
13

Is this statement true or false? Gymnosperms reproduce using seeds, but angiosperms do not.

Chemistry
1 answer:
spayn [35]3 years ago
3 0

Answer:

FALSE!

Explanation:

Gymnosperms do use seeds but are exposed like the pine cones of pines. Angiosperms still have seeds, however, they flower or fruit.

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Chuẩn Độ 15 ml dung dịch CH3COOH 0,2 m bằng dung dịch NaOH 0,2 m
mina [271]

Answer:

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Explanation:

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3 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.

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3 years ago
Which dilute acid is used to make sodium carbonate salt ​
andreev551 [17]
Hydrochloric acid :))
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The reaction described in Part A required 3.54 L of magnesium chloride. What is the concentration of this magnesium chloride sol
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we need the cholride solution Part B to answer your question

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