Answer:
C
Explanation:
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<u>Answer:</u> The pH of the buffer is 4.61
<u>Explanation:</u>
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
We are given:
= negative logarithm of acid dissociation constant of weak acid = 4.70
= moles of conjugate base = 3.25 moles
= Moles of acid = 4.00 moles
pH = ?
Putting values in above equation, we get:
Hence, the pH of the buffer is 4.61
In a bag of peas that weighs 454 grams, there are between 1261 and 4540 peas.
The average pea weighs between 0.1 and 0.36 grams.
If we take the lower value (0.1 g/pea), the number of peas in 454 g is:
If we take the higher value (0.36 g/pea), the number of peas in 454 g is:
In a bag of peas that weighs 454 grams, there are between 1261 and 4540 peas.
You can learn more about conversion factors here: brainly.com/question/1844638
The reaction between copper II chloride and sodium sulfide as well as lead II nitrate and potassium sulfate both produce precipitates.
The solubility of a substance in water is in accordance with the solubility rules. It is possible that a solid product may be formed when two aqueous solutions are mixed together. That solid product is referred to as a precipitate.
Now, we will consider each reaction individually to decode whether or not a precipitate is possible.
- In the first reaction, we have; CuCl2(aq) + Na2S(aq) ---->CuS(s) + 2NaCl(aq). A precipitate (CuS) is formed.
- In the second reaction, Pb(NO3)2(aq) + 2KNO3(aq) ----> PbSO4(s) + KNO3(aq), a precipitate PbSO4 is formed
- In the third reaction, NH4Br(aq) + NaOH(aq) ----->NH3(g) + NaBr(aq) + H2O(l), a precipitate is not formed here.
Learn more: brainly.com/question/11969651