Answer:
Explanation:
The usefulness of a buffer is its ability to resist changes in pH when small quantities of base or acid are added to it. This ability is the consequence of having both the conjugate base and the weak acid present in solution which will consume the added base or acid.
This capacity is lost if the ratio of the concentration of conjugate base to the concentration of weak acid differ by an order of magnitude. Since buffers having ratios differing by more will have their pH driven by either the weak acid or its conjugate base .
From the Henderson-Hasselbach equation we have that
pH = pKa + log [A⁻]/[HA]
thus
0.1 ≤ [A⁻]/[HA] ≤ 10
Therefore the log of this range is -1 to 1, and the pH will have a useful range of within +/- 1 the pKa of the buffer.
Now we are equipped to answer our question:
pH range = 3.9 +/- 1 = 2.9 through 4.9
Answer:
3 NH4OH (l) + H3PO4 (aq) → (NH4)3PO4 (aq) + 3 H2O (l)
Explanation:
This is an acid-base reaction (neutralization): NH4OH is a base, H3PO4 is an acid
2 atome nitrogen , 1 is correct
Answer:
It would take 3.11 J to warm 3.11 grams of gold
Explanation:
Step 1: Data given
Mass of gold = 3.11 grams
Temperature rise = 7.7 °C
Specific heat capacity of gold = 0.130 J/g°C
Step 2: Calculate the amount of energy
Q = m*c*ΔT
⇒ Q = the energy required (in Joules) = TO BE DETERMINED
⇒ m = the mass of gold = 3.11 grams
⇒ c = the specific heat of gold = 0.130 J/g°C
⇒ ΔT = The temperature rise = 7.7 °C
Q = 3.11 g * 0.130 J/g°C * 7.7 °C
Q = 3.11 J
It would take 3.11 J to warm 3.11 grams of gold
.. it’s the answer 0.25 g MgSO4? ..
I may be wrong.
