Hello!
First, we can assume the density of water to be
1 g/mL, so the volume of water would be
725 mL. The density of Acetic Acid (Pure) is
1,05 g/mL so 8 grams would represent
7,61 mLNow we can apply the following conversion factor to calculate the molarity of the solution, using the molar mass of Acetic Acid:
![[CH_3CO_2H]=\frac{8 g_{CH_3CO_2H} }{725 mL + 7,61 mL}* \frac{1000 mL}{1 L}* \frac{1 mol_{CH_3CO_2H} }{60,05 g_{CH_3CO_2H} }](https://tex.z-dn.net/?f=%5BCH_3CO_2H%5D%3D%5Cfrac%7B8%20g_%7BCH_3CO_2H%7D%20%20%7D%7B725%20mL%20%2B%207%2C61%20mL%7D%2A%20%5Cfrac%7B1000%20mL%7D%7B1%20L%7D%2A%20%5Cfrac%7B1%20mol_%7BCH_3CO_2H%7D%20%7D%7B60%2C05%20g_%7BCH_3CO_2H%7D%20%7D)
![[CH_3CO_2H]=0,1818 M](https://tex.z-dn.net/?f=%5BCH_3CO_2H%5D%3D0%2C1818%20M)
So, the concentration of acetic acid would be
0,1818 MHave a nice day!
The answer is A. Increasing temperature of the reactants
Answer: 44.37 degrees C
Explanation:
Use combined gas law: (P1)(V1)/T1=(P2)(V2)/T2
For most gas laws, you must convert to Kelvin:
K=deg C+273
K=25+273=298 K
Plug and chug:
(1.0 atm)(1.2 L)/(298 K)=(0.71 atm)(1.8 L)/(x)
Solve for x and get 317.37 K
Subtract 273 from this to convert to degrees Celsius. You will get 44.37 degrees Celsius.
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Answer:
A
Explanation:
You want an equation that includes both V and T. Charles' Law states that V =kT or V₁/T₁ = V₂/T₂, so this is the best formula to use.
B: The Combined Gas Law is p₁V₁/T₂ = p₂V₂/T₂ will work, but it's overkill for the situation. You can assume any constant value for the pressure, and it will cancel from each side of the equation,
C is wrong. Boyle's Law is p₁V₁ = p₂V₂. It does not include the temperature.
D is wrong. Gay-Lussac's Law is p₁/T₂ = p₂/T₂. It does not include the volume.