Answer:
need more info
Explanation:
i think so we need more info
Answer:
The student measured the cup and got an incorrect answer
Explanation:
because he was in a rush.
Answer:
0.056moles HF and 0.70M
Explanation:
When a strong acid is added to a buffer, the acid reacts with the conjugate base.
In the system, NaF and HF, weak acid is HF and conjugate base is NaF. The reaction of NaF with HCl (Strong acid) is:
NaF + HCl → HF + NaCl
Initial moles of NaF and HF in 60.0mL of solution are:
NaF:
0.0600L × (0.80mol / L)= 0.048 moles NaF
HF:
0.0600L × (0.80mol / L)= 0.048 moles HF
Then, the added moles of HCl are:
0.0200L × (0.40mol / L) = 0.008 moles HCl.
Thus, after the reaction, moles of HF produced are 0.008 moles + the initial 0.048moles of HF, moles of HF are:
<em>0.056moles HF</em>
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In 20.0mL + 60.0mL = 80.0mL = 0.0800L, molarity of HF is:
0.056mol HF / 0.0800L = <em>0.70M</em>
The balanced equation is Mg(s) + 2HCI(aq) = H2(g) + MgCI2 (aq)
The percentage of glucose given is m/v. This means that the given percentage of volume consists of mass.
In this solution, percentage of glucose is 5.5% m/v.
This means that 5.5% of the volume is the mass of glucose.
Given volume is 285 mL.
Therefore mass of glucose is 5.50% of 285 mL = (5.5*285)/100
mass of glucose = 15.67 g
