Buffers - mixtures of conjugate acid and conjugate base at ±1 pH unit from pH = pKa. Resistant to changes in pH in response to small additions of H+ or OH-. ... Polyprotic acids - dissociation of each H+ can be treated separately if the pKa values are different
Although the models are not provided, I was able to find them and the beakers with solid present in them are:
1C
2A
2C
3A
3C
This is determined by the fact that the beakers all have a piece of closely packed substance laying at the bottom. This closely packed lattice is characteristic of solid substances, and the fact that they exist in the solution in the solid states indicates that they are insoluble.
Answer:
the Molar heat of Combustion of diphenylacetylene
= 
Explanation:
Given that:
mass of diphenylacetylene
= 0.5297 g
Molar Mass of diphenylacetylene
= 178.21 g/mol
Then number of moles of diphenylacetylene
= 
= 
= 0.002972 mol
By applying the law of calorimeter;
Heat liberated by 0.002972 mole of diphenylacetylene
= Heat absorbed by
+ Heat absorbed by the calorimeter
Heat liberated by 0.002972 mole of diphenylacetylene
= msΔT + cΔT
= 1369 g × 4.184 J g⁻¹°C⁻¹ × (26.05 - 22.95)°C + 916.9 J/°C (26.05 - 22.95)°C
= 17756.48 J + 2842.39 J
= 20598.87 J
Heat liberated by 0.002972 mole of diphenylacetylene
= 20598.87 J
Heat liberated by 1 mole of diphenylacetylene
will be = 
= 6930979.139 J/mol
= 6930.98 kJ/mol
Since heat is liberated ; Then, the Molar heat of Combustion of diphenylacetylene
= 
Charles law gives the relationship between volume and temperature of gas.
It states that at constant pressure volume is directly proportional to temperature
Therefore
V/ T = k
Where V - volume T - temperature in kelvin and k - constant
V1/T1 = V2/T2
Parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
Substituting the values in the equation
267 L/ 480 K = V / 750 K
V = 417 L
Final volume is 417 L
Answer:
25 mL
Explanation:
Step 1: Given data
- Concentration of the concentrated solution (C₁): 2 M
- Volume of the concentrated solution (V₁): ?
- Concentration of the diluted solution (C₂): 0.1 M
- Volume of the diluted solution (V₂): 0.500 L
Step 2: Calculate the volume of the concentrated NaCl solution
We will use the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂ / C₁
V₁ = 0.1 M × 0.500 L / 2 M
V₁ = 0.025 L = 25 mL