<span>Heat
gained or absorbed in a system can be calculated by multiplying the given mass to the
specific heat capacity of the substance and the temperature difference. The heat capacity of aluminum at 25 degrees celsius is 0.9 J/g-C. It is
expressed as follows:</span><span>
Heat = mC(T2-T1)
5800 J = 152000(0.90)(</span>ΔT)
ΔT = 0.42 °C change in temperature
Answer:
C3H6O2
Explanation:
To find the empirical formula of the compound, we divide the amount in moles of each of the elements by the amount in mole of the element with the smallest number of mole. In this question, the element with the smallest number of moles is oxygen with 1.36 mole. Hence, we divide the number of moles of each element by this.
H = 4.10/1.36 = 3
O = 1.36/1.36 = 1
C = 2.05/1.36 = 1.5
We then multiply through by 2 to yield the compound with the empirical formula C3H6O2
Answer: 2-Phenyl-2-Penetene on ozonolysis <span>yields acetophenone and propanal.
Explanation: The tricky way to solve such questions is to simply break the double bond in alkene and place oxygen atom at each broken half double bond making it carbonyl group. The reaction of given statement is as follow,</span>
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
Answer:
1.98 M
Explanation:
Given data
- Initial volume (V₁): 93.2 mL
- Initial concentration (C₁): 2.03 M
- Volume of water added: 3.92 L
Step 1: Convert V₁ to liters
We will use the relationship 1 L = 1000 mL.
Step 2: Calculate the final volume (V₂)
The final volume is the sum of the initial volume and the volume of water.
Step 3: Calculate the final concentration (C₂)
We will use the dilution rule.
