Answer:
0,07448M of phosphate buffer
Explanation:
sodium monohydrogenphosphate (Na₂HP) and sodium dihydrogenphosphate (NaH₂P) react with HCl thus:
Na₂HP + HCl ⇄ NaH₂P + NaCl <em>(1)</em>
NaH₂P + HCl ⇄ H₃P + NaCl <em>(2)</em>
The first endpoint is due the reaction (1), When all phosphate buffer is as NaH₂P form, begins the second reaction. That means that the second endpoint is due the total concentration of phosphate that is obtained thus:
0,01862L of HCl×
= 1,862x10⁻³moles of HCl ≡ moles of phosphate buffer.
The concentration is:
= <em>0,07448M of phosphate buffer</em>
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Answer:
D. cathode; reduction
Explanation:
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Answer:
1.714 M
Explanation:
We'll begin by calculating the number of mole in 46.8 g of NaHCO₃. This can be obtained as follow:
Mass of NaHCO₃ = 46.8 g
Molar mass of NaHCO₃ = 23 + 1 + 12 + (3×16)
= 23 + 1 + 12 + 48
= 84 g/mol
Mole of NaHCO₃ =?
Mole = mass / molar mass
Mole of NaHCO₃ = 46.8 / 84
Mole of NaHCO₃ = 0.557 mole
Next, we shall convert 325 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
325 mL = 325 mL × 1 L / 1000 mL
325 mL = 0.325 L
Thus, 325 mL is equivalent to 0.325 L.
Finally, we shall determine the molarity of the solution. This can be obtained as shown below:
Mole of NaHCO₃ = 0.557 mole
Volume = 0.325 L
Molarity =?
Molarity = mole / Volume
Molarity = 0.557 / 0.325
Molarity = 1.714 M
Therefore the molarity of the solution is 1.714 M
Answer:
A change that produces one or more new substances is a chemical change or chemical reaction.
