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
Answer: 45.038
Explanation: 2.016 H (2*1.008) + 15.999 O (1*15.999)
88.98 %
The Balance Chemical Equation is as follow,
2 HCl + Pb(NO₃)₂ → 2 HNO₃ + PbCl₂
According to equation,
331.2 g (1 mole) Pb(NO₃)₂ produces = 278.1 g (1 mole) PbCl₂
So,
870 g of Pb(NO₃)₂ will produce = X g of PbCl₂
Solving for X,
X = (870 g × 278.1 g) ÷ 331.2 g
X = 730.5 g of PbCl₂
Therefore,
Theoretical Yield = 730.5 g
Also as given,
Actual Yield = 650 g
So using following formula for percentage yield,
%age Yield = (Actual Yield / Theoretical Yield) × 100
Putting values,
%age Yield = (650 g / 730.5 g) × 100
%age Yield = 88.98 %
Brianliest please and thank you.
They are the same because the definition of a mole is 6.022 x 10^23 molecules or atoms based on whether it is a molecule or element. so there are 6.022 x 10^23 atoms of argon in one mole of argon and 6.022 x 10^23 molecules of ammonia in one mole of ammonia
This is a PV=nRT problem. So to do this, you are looking for grams. The easiest way to get to grams is to convert moles, so set up the equation looking for n
n=PV/RT
To keep everything consistent with the universal gas constant (0.082057 L atm/mol K), one thing needs to be converted.
To convert 34.0 C to kelvin, just add 273.15. That gets 307.2 with sig figs.
Next, plug everything into your equation
n=(6.22atm)(16.6L)/(0.082057Latm/molK)(307.2K)
Cleaning that up gets you 4.10 moles (again, after sig figs)
Now, you need to convert the 4.10 moles of O2 to grams. To do that, you multiply 4.10 by 31.998 (molar mass of o2) to get 131 g (for a third time, sig figs!)
So, your final answer is 131 g O2
Sorry if any math is wrong, but you should be able to walk yourself through this and other problems now. Hope I helped!
