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: The correct option is 2.
Explanation: Heat flow is defined as the transfer of energy from hotter object to cooler object when two objects are kept together at different temperatures. As the energy remains conserved, so the heat flow will take place until the equilibrium is attained.
In the above asked question, Object A is at 40° C and Object B is at 80° C.
Object B is at higher temperature, so the heat flow will take place from Object B to Object A.
Hence, the correct option is 2.
<u>Answer:</u> The correct answer is Option A.
<u>Explanation:</u>
Electronegativity is defined as the tendency of an atom to attract the shared pair of electrons towards itself whenever a bond is formed.
This property increases as we move from left to right across a period because the number of charge on the nucleus gets increased and electrons are attracted more towards the nucleus.
This property decreases as we move from top to bottom in a group because the electrons get add up in the new shells which make them further away from the nucleus.
Thus, the correct answer is Option A.
Answer:
C. 0.4.
Explanation:
<em>∵ mole fraction of acetic acid (X acetic acid) = (no. of moles acetic acid)/(total no. of moles) = (no. of moles acetic acid)/(no. of moles of acetic acid + no. of moles of water).</em>
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- no. of moles of acetic acid = 2, no. of moles of water = 3.
- Total no. of moles = no. of moles of acetic acid + no. of moles of water = 2 + 3 = 5.
<em>∴ mole fraction of acetic acid (X acetic acid) = (no. of moles acetic acid)/(total no. of moles) =</em> (2)/(5)<em> = 0.4.</em>
