Answer:
2.448 * 10^-5M
Explanation:
To calculate the concentration of the solution, we apply the Beer Lambert law equation.
That is A = £LC
Where £ represents the molar extinction coefficient.
We identify the values as follows:
A = absorbance = 0.735
L = cell length = 1.30
£ = 23100
c = concentration = ?
Rearranging the equation, c = A/£L
c = 0.735/(23,100 * 1.3)
c = 2.448 * 10^-5 M
Answer:
37064 J
Explanation:
Data Given:
mass of Steam (m) = 16.4 g
heat released (Q) = ?
Solution:
This question is related to the latent heat of condensation.
Latent heat of condensation is the amount of heat released when water vapors condenses to liquid.
Formula used
Q = m x Lc . . . . . (1)
where
Lc = specific latent heat of condensation
Latent heat of vaporization of water is exactly equal to heat of condensation with - charge
So, Latent heat of vaporization of water have a constant value
Latent heat of vaporization of water = 2260 J/g
So
Latent heat of condensation of water will be = - 2260 J/g
Put values in eq. 1
Q = (16.4 g) x (- 2260 J/g)
Q = - 37064 J
So, 37064 J of heat will be released negative sign indicate release of energy
<h2>Gas Molecules Move in Random Motion </h2>
Explanation:
- According to the property of an ideal gas the volume occupied by the gas molecules themselves is negligible as compared to the volume occupied by the gas
- The molecules of ideal gas obey Newton's laws of motion thus they move in random motion
- According to the above statements we can conclude that Helium gas can fill many more balloons as compared to fit the baloons inside the cylinder because the particles of gas is free to move in random motion and can occupy more volume
Answer:
pH = 10.9
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to say that the undergoing reaction between this buffer and OH⁻ promotes the formation of more CO₃²⁻ because it acts as the base, we can do the following:
The resulting concentrations are:
![[CO_3^{2-}]=\frac{0.1435mol}{0.25L}=0.574M \\](https://tex.z-dn.net/?f=%5BCO_3%5E%7B2-%7D%5D%3D%5Cfrac%7B0.1435mol%7D%7B0.25L%7D%3D0.574M%20%5C%5C)
![[HCO_3^{-}]=\frac{0.0265mol}{0.25L}=0.106M](https://tex.z-dn.net/?f=%5BHCO_3%5E%7B-%7D%5D%3D%5Cfrac%7B0.0265mol%7D%7B0.25L%7D%3D0.106M)
Thus, since the pKa of this buffer system is 10.2, the change in the pH would be:
Which makes sense since basic OH⁻ ions were added.
Regards!
