Answer:
92.9 °C
Explanation:
Step 1: Given data
- Initial volume (V₁): 450. mL
- Initial temperature (T₁): 55.0 °C
- Final volume (V₂): 502 mL
Step 2: Convert 55.0 °C to Kelvin
We will use the following expression.
K = °C + 273.15 = 55.0 + 273.15 = 328.2 K
Step 3: Calculate the final temperature of the gas
If we assume constant pressure and ideal behavior, we can calculate the final temperature of the gas using Charles' law.
T₁/V₁ = T₂/V₂
T₂ = T₁ × V₂/V₁
T₂ = 328.2 K × 502 mL/450. mL = 366 K = 92.9 °C
Here, we should use combined gas law which can be derived from combined gas law, “PV=nRT”. Rearranging, we can get PV/T=nR. Then we can set the two states in the problem together to get
P1V1/T1 = P2V2/T2
Then just plug in and solve algebraically.
Hope this helps
A special type of combination is combustion. One of the key reactants in a combustion reaction must always involve an oxygen. This is the defining characteristic of combustion. When a combustible matter is burned, it reacts with the oxygen in air in the presence of heat. Moreover, one of its end products must be water. Let's take glucose as our example of a hydrocarbon.
C6H12O6(s) + 6O2(g) -----> 6CO2 (g) + H2O(g)
Therefore, when a solid glucose interacts with oxygen gas, it yields carbon dioxide and water vapor.
Answer:
Explanation:
The limiting reactant is the reactant that gives the smaller amount of product.
Assemble all the data in one place, with molar masses above the formulas and masses below them.
M_r: 39.10 80.41 2.016
2K + 2HBr ⟶ 2KBr + H₂
m/g: 5.5 4.04
a) Limiting reactant
(i) Calculate the moles of each reactant
(ii) Calculate the moles of H₂ we can obtain from each reactant.
From K:
The molar ratio of H₂:K is 1:2.
From HBr:
The molar ratio of H₂:HBr is 3:2.
(iii) Identify the limiting reactant
HBr is the limiting reactant because it gives the smaller amount of NH₃.
b) Excess reactant
The excess reactant is K.
c) Mass of H₂
