<span>Answer:
(Ka = 6.8*10^-4). K3PO4 is the salt of a weak acid and a strong base. It ionizes mostly, (Kb = 0.025) and is a mild electrolyte.</span>
Answer:
Dipole-dipole attractions
Explanation:
The Cl atom is more electronegative than H, so it attracts the electrons in the H-Cl bond more to itself.
The Cl end of the molecule becomes partially negative (δ-), and the H end becomes more positive (δ+). This separation of charge is called an electric dipole,
When two HCl molecules are near each other, they arrange themselves so that the negative and positive ends of the dipoles line up and attract the two molecules together.
Answer:
One can conclude that the metal is an alloy.
Explanation:
An alloy is a combination of metal that has two or more elements. Different metals are usually combined to give it more strength or make it more resistant to corrosion.
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Answer:
Percent yield = 94.5%
Theoretical yield = 26.89 g
Explanation:
Given data:
Mass of copper = 12.5 g
Mass of copper chloride produced = 25.4 g
Theoretical yield = ?
Percent yield = ?
Solution:
Cu + Cl₂ → CuCl₂
Number of moles of Copper:
Number of moles = mass/ molar mass
Number of moles = 12.5 g/ 63.55 g/mol
Number of moles = 0.2 mol
Now we will compare the moles of copper with copper chloride.
Cu : CuCl₂
1 : 1
0.2 : 0.2
Theoretical yield:
Mass of copper chloride:
Mass = Number of moles × molar mass
Mass = 0.2 mol × 134.45 g/mol
Mass = 26.89 g
Percent yield:
Percent yield = Actual yield / theoretical yield × 100
Percent yield = 25.4 g/26.89 g × 100
Percent yield = 94.5%
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>