For all three questions, we will use the fact that
- molarity = (moles of solute)/(liters of solution)
1) For 175 mL of solution at 0.203 M, this means that:
- 0.203 = (moles of solute)/0.175
- moles of solute = 0.035523 mol
Considering the hydrochloric acid solution, if we have 0.035523 mol, then:
- 6.00 = 0.035523/(liters of solution)
- liters of solution = 0.035523/6.00 = 0.0059205 = <u>5.92 mL (to 3 sf)</u>
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2) If there is 20.3 mL = 0.0203 L, then:
- 8.20 = (moles of solute)/0.0203
- moles of solute = 0.16646 mol
This means that the molarity of the diluted solution is:
- 0.16646/(0.200) = <u>0.832 M (to 3 sf)</u>
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3) If we need 1.50 L of 0.700 M solution, then:
- 0.700 = (moles of solute)/1.50
- moles of solute = 1.05 mol
Considering the 9.36 M acid solution, from which we need 1.05 mol of perchloric acid from,
- 9.36 = 1.05/(liters of solution)
- liters of solution = 1.05/9.36, which is 0.11217948717949 L, or <u>112 mL (to 3 sf)</u>
44.0095 you're welcome hope this helps
What are the sentences?
It would help to see the sentences.
Bohr suggested, that there are definitive shells of particular energy and angular momentum in which an electron can revolve. It was not in Rutherford's model
Explanation:
Since HF is a weak acid, the use of an ICE table is required to find the pH. The question gives us the concentration of the HF.
HF+H2O⇌H3O++F−HF+H2O⇌H3O++F−
Initial0.3 M-0 M0 MChange- X-+ X+XEquilibrium0.3 - X-X MX M
Writing the information from the ICE Table in Equation form yields
6.6×10−4=x20.3−x6.6×10−4=x20.3−x
Manipulating the equation to get everything on one side yields
0=x2+6.6×10−4x−1.98×10−40=x2+6.6×10−4x−1.98×10−4
Now this information is plugged into the quadratic formula to give
x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)−−−−−−−−−−−−−−−−−−−−−−−−−−−−√2x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)2
The quadratic formula yields that x=0.013745 and x=-0.014405
However we can rule out x=-0.014405 because there cannot be negative concentrations. Therefore to get the pH we plug the concentration of H3O+ into the equation pH=-log(0.013745) and get pH=1.86
