Answer:
They can both be used in an electric circuit.
Explanation:
Though conductors and insulators are very different from each other but still they have a similarity that "they can both be used in an electric circuit".
Conductors are the material that allows electricity to flow and insulators have high resistance and do not allow electricity to flow.
Both conductors and insulators are used in an electric circuit as conductor is used in wires, batteries and bulb to flow current while insulators are used in insulation of wire, switches, plugs and etc. Combination of them makes the circuit shock free.
Hence, the correct answer is "They can both be used in an electric circuit."
If the actual yield of sodium chloride from the reaction of 8.3 g of sodium and 4.5 g of chlorine is 6.4 g, what is the percent yield?
Answer : 2Na(s) + Cl2(g) → 2NaCl(s)
Explanation: Friend me!!!
Answer:
2.25g of NaF are needed to prepare the buffer of pH = 3.2
Explanation:
The mixture of a weak acid (HF) with its conjugate base (NaF), produce a buffer. To find the pH of a buffer we must use H-H equation:
pH = pKa + log [A-] / [HA]
<em>Where pH is the pH of the buffer that you want = 3.2, pKa is the pKa of HF = 3.17, and [] could be taken as the moles of A-, the conjugate base (NaF) and the weak acid, HA, (HF). </em>
The moles of HF are:
500mL = 0.500L * (0.100mol/L) = 0.0500 moles HF
Replacing:
3.2 = 3.17 + log [A-] / [0.0500moles]
0.03 = log [A-] / [0.0500moles]
1.017152 = [A-] / [0.0500moles]
[A-] = 0.0500mol * 1.017152
[A-] = 0.0536 moles NaF
The mass could be obtained using the molar mass of NaF (41.99g/mol):
0.0536 moles NaF * (41.99g/mol) =
<h3>2.25g of NaF are needed to prepare the buffer of pH = 3.2</h3>
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