Answer:
![[base]=0.28M](https://tex.z-dn.net/?f=%5Bbase%5D%3D0.28M)
Explanation:
Hello,
In this case, by using the Henderson-Hasselbach equation one can compute the concentration of acetate, which acts as the base, as shown below:
![pH=pKa+log(\frac{[base]}{[acid]} )\\\\\frac{[base]}{[acid]}=10^{pH-pKa}\\\\\frac{[base]}{[acid]}=10^{4.9-4.76}\\\\\frac{[base]}{[acid]}=1.38\\\\](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29%5C%5C%5C%5C%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%3D10%5E%7BpH-pKa%7D%5C%5C%5C%5C%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%3D10%5E%7B4.9-4.76%7D%5C%5C%5C%5C%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%3D1.38%5C%5C%5C%5C)
![[base]=1.38[acid]=1.38*0.20M=0.28M](https://tex.z-dn.net/?f=%5Bbase%5D%3D1.38%5Bacid%5D%3D1.38%2A0.20M%3D0.28M)
Regards.
Given:
Ma = 31.1 g, the mass of gold
Ta = 69.3 °C, the initial temperature of gold
Mw = 64.2 g, the mass of water
Tw = 27.8 °C, the initial temperature of water
Because the container is insulated, no heat is lost to the surroundings.
Let T °C be the final temperature.
From tables, obtain
Ca = 0.129 J/(g-°C), the specific heat of gold
Cw = 4.18 J/(g-°C), the specific heat of water
At equilibrium, heat lost by the gold - heat gained by the water.
Heat lost by the gold is
Qa = Ma*Ca*(T - Ta)
= (31.1 g)*(0.129 J/(g-°C)(*(69.3 - T °C)-
= 4.0119(69.3 - T) j
Heat gained by the water is
Qw = Mw*Cw*(T-Tw)
= (64.2 g)*(4.18 J/(g-°C))*(T - 27.8 °C)
= 268.356(T - 27.8)
Equate Qa and Qw.
268.356(T - 27.8) = 4.0119(69.3 - T)
272.3679T = 7738.32
T = 28.41 °C
Answer: 28.4 °C
CH₇ is the empirical formula of the car fuel.
Explanation:
To find the empirical formula we use the following algorithm.
First divide each mass the the molar weight of each element:
for carbon 2.87 / 12 = 0.239
for hydrogen 3.41 / 2 = 1.705
And now divide each quantity by the lowest number which is 0.239:
for carbon 0.239 / 0.239 = 1
for hydrogen 1.705 / 0.239 = 7.13 ≈ 7
The empirical formula of the car fuel is CH₇.
I have to tell you that in reality this formula is wrong because is not possible to exist. However the algorithm for finding the empirical formula is right, the problem may reside in the amounts of carbon and hydrogen given.
Learn more about:
empirical formula
brainly.com/question/5297213
#learnwithBrainly
Answer:
8.934 g
Step-by-step explanation:
We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 192.12 44.01
H₃C₆H₅O₇ + 3NaHCO₃ ⟶ Na₃C₆H₅O₇ + 3H₂O + 3CO₂
m/g: 13.00
For ease of writing, let's write H₃C₆H₅O₇ as H₃Cit.
(a) Calculate the <em>moles of H₃Cit
</em>
n = 13.00 g × (1 mol H₃Cit /192.12 g H₃Cit)
n = 0.067 67 mol H₃Cit
(b) Calculate the <em>moles of CO₂
</em>
The molar ratio is (3 mol CO₂/1 mol H₃Cit)
n = 0.067 67 mol H₃Cit × (3 mol CO₂/1 mol H₃Cit)
n = 0.2030 mol CO₂
(c) Calculate the <em>mass of CO₂
</em>
m = 0.2030 mol CO₂ × (44.01 g CO₂/1 mol CO₂)
m = 8.934 g CO₂
