Answer: A 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with 3 mL volume.
Explanation:
Given: 
= 0.20 M, 
= 15.0 mL
= 0.10 M, 
= ?
Formula used is as follows.
Substitute the values into above formula s follows.
![M_{1}V_{1} = M_{2}V_{2}\\0.20 M ]times 15.0 mL = 0.10 M ]times V_{2}\\V_{2} = 30 mL](https://tex.z-dn.net/?f=M_%7B1%7DV_%7B1%7D%20%3D%20M_%7B2%7DV_%7B2%7D%5C%5C0.20%20M%20%5Dtimes%2015.0%20mL%20%3D%200.10%20M%20%5Dtimes%20V_%7B2%7D%5C%5CV_%7B2%7D%20%3D%2030%20mL)
Thus, we can conclude that a 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with 3 mL volume.
Answer:
The correct answer to the question which statements about Avogadro’s Law is false is
c. At constant T and P, doubling the moles of gas decreases the volume by half.
Explanation:
Avogadro's law describes the relationship between the volume of a mass of gas and the number of moles present. Avogadro's law states that at standard (or the same) temperature and pressure, equal volumes od all gases contain equal number of molecules
That is mathematically
where
V₁ = volume of first sample
V₂ = volume of second sample
n₁ = number of moles in first sample
n₂ = number of moles in second sample
<span>Answer: 17.8 cm
</span>
<span>Explanation:
</span>
<span>1) Since temperature is constant, you use Boyle's law:
</span>
<span>PV = constant => P₁V₁ = P₂V₂
</span><span>=> V₁/V₂ = P₂/P₁</span>
<span>
2) Since the ballon is spherical:
</span><span>V = (4/3)π(r)³</span>
<span>
Therefore, V₁/V₂ = (r₁)³ / (r₂)³
</span>
<span>3) Replacing in the equation V₁/V₂ = P₂/P₁:
</span><span><span>(r₁)³ / (r₂)³ </span>= P₂/P₁</span>
<span>
And you can solve for r₂: (r₂)³ = (P₁/P₂) x (r₁)³
</span>(r₂)³ = (1.0 atm / 0.87 atm) x (17 cm)³ = 5,647.13 cm³
<span>
r₂ = 17.8 cm</span>
Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
Heat capacity of the bomb plus water = 10.4 KJ/oC
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
qcal = (10.4 kJ/°C)(4.20°C)
= 43.68 kJ
Always qsys = qcal + qrxn = 0,
qrxn = -43.68 kJ
The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is:
