i believe the answer is because asphalt conducts heat better
Answer : The mass of sodium acetate is, 1.097 grams.
Explanation : Given,
The dissociation constant for acetic acid = 
Concentration of acetic acid (weak acid)= 0.20 M
volume of solution = 125. mL
pH = 4.47
First we have to calculate the value of 
.
The expression used for the calculation of is,
Now put the value of 
in this expression, we get:
Now we have to calculate the concentration of sodium acetate (conjugate base or salt).
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
Now put all the given values in this expression, we get:
![4.47=4.74+\log (\frac{[Salt]}{0.20})](https://tex.z-dn.net/?f=4.47%3D4.74%2B%5Clog%20%28%5Cfrac%7B%5BSalt%5D%7D%7B0.20%7D%29)
![[Salt]=0.107M](https://tex.z-dn.net/?f=%5BSalt%5D%3D0.107M)
Now we have to calculate the mass of sodium acetate.
Therefore, the mass of sodium acetate is, 1.097 grams.
Which.i think its the answer
Answer:
0.500 moles of CO2 has a volume of 11.2 L at STP (option B)
Explanation:
Step 1: Data given
Volume of a gas at STP = 11.2 L
STP: Pressure = 1 atm and temperature = 273 K
Step 2: Calculate volume
p*V= n*R*T
V = (n*R*T)/p
⇒with V = the volume of the gas = TO BE DETERMINED
⇒with n = the number of moles of the gas
⇒with R = the gas constant = 0.08206 L*atm/mol*K
⇒with T = the temperature = 273 K
⇒with p = the pressure of the gas = 1 atm
A
) 0.250 mole of NH3
V = (0.250 * 0.08206 * 273) / 1
V = 5.6 L
B
) 0.500 mole of CO2
V = (0.500 * 0.08206 * 273) / 1
V = 11.2 L
C
) 0.750 mole of NH3
V = (0.750 * 0.08206 * 273) / 1
V = 16.8 L
D) 1.00 mole of CO2
V = (1.00 * 0.08206* 273) / 1
V = 22.4 L
0.500 moles of CO2 has a volume of 11.2 L at STP (option B)
<span>d, The books have a great deal of inertia and do not move easily.</span>
