Answer:
0.210 M
Explanation:
<em>A 75.0 mL aliquot of a 1.70 M solution is diluted to a total volume of 278 mL.</em>
In order to find out the resulting concentration (C₂) we will use the dilution rule.
C₁ × V₁ = C₂ × V₂
1.70 M × 75.0 mL = C₂ × 278 mL
C₂ = 0.459 M
<em>A 139 mL portion of that solution is diluted by adding 165 mL of water. What is the final concentration? Assume the volumes are additive.</em>
Since the volumes are additive, the final volume V₂ is 139 mL + 165 mL = 304 mL. Next, we can use the dilution rule.
C₁ × V₁ = C₂ × V₂
0.459 M × 139 mL = C₂ × 304 mL
C₂ = 0.210 M
The reaction is : C3H8 + 5O2 ===> 3CO2 + 4H2O
find the number of moles of H2O using the molar mass of H2O
moles of H2O = 75g/18.02 g/mol = 4.162 moles
go between the H2O and the propane to find the mole of propane
4.162 moles H2O its a 1 : 4 ratio so divide 4.162 / 4 = 1.041mol of propane
use this equation to find the volume of gas required : V = nRT/P
n is the moles of gas (propane)
R is the universal gas constant (0.082057338)
T is the temperature of the gas (273K)
V = (1.041) x (0.08206 L-atm/mol-K) x (273K)/(1 atm)
V = 23.3 L of C3H8
hope that helps
As the temperature of a liquid increases, its viscosity decreases.
Answer:
C
Explanation:
This is a correct answer i think soo
Hello!
To solve this question, we first have to know the
pKa of the acid:
Now, we can apply the
Henderson-Hasselbach's equation to determine the pH of the buffer solution:
![pH=pKa+ log( \frac{[NaA]}{[HA]} )= 6,25 + log(\frac{(0,609 mol/2L)}{(0,506 mol/2L)})=6,33](https://tex.z-dn.net/?f=pH%3DpKa%2B%20log%28%20%5Cfrac%7B%5BNaA%5D%7D%7B%5BHA%5D%7D%20%29%3D%206%2C25%20%2B%20log%28%5Cfrac%7B%280%2C609%20mol%2F2L%29%7D%7B%280%2C506%20mol%2F2L%29%7D%29%3D6%2C33%20)
So, the
pH of this buffer solution is 6,33Have a nice day!
