6.5-7.5 i believe was right
The total percent yield:
After the combustion reaction with methane, the percent yield was 66.7%.
Combustion of Methane:
- Methane produces a blue flame as it burns in the atmosphere.
- Methane burns in the presence of enough oxygen to produce carbon dioxide (CO₂) and water (H₂O).
- It creates a significant quantity of heat during combustion, making it an excellent fuel source.
The other reactant, air's excess oxygen, is always present, making methane the limiting reactant. As a result, the amount of CH₄ burned will determine how much CO₂ and H₂O are produced.
The following chemical process produces carbon dioxide from methane:
CH₄ + 2O₂ ⇒ CO₂ + 2H₂O
Calculations:
1. <u><em>Theoretical quantity of carbon dioxide:</em></u>
All calculations will be based on the amount of methane because the problem specifies that it is the limiting reagent:
12.0g of CH₄ × (1 mol of CH₄/16g CH₄) × (1 mole of CO₂/1 mole of CH₄) × (44g CO₂/1 mole of CO₂)
= 33g of CO₂
2. <u><em>Percent yield:</em></u>
= Actual yield/Theoretical yield × 100
= 22.0g/33g × 100
= 66.7%
Learn more about the percent yield here,
brainly.com/question/15535037
#SPJ4
Answer:
Zn2+ is colourless
Explanation:
We know that transition metal salts are usually coloured due to the possibility of d-d transition.
This d-d transition can only occur when there are vacant d-orbitals. The electronic configuration [Ar] 4s23d8 suggests the presence of vacant d-orbitals and the possibility of the compounds of Zn2+ being coloured.
However, the absence of colours in Zn2+ compounds shows that there is no d-d transition(electronic) spectra observed for Zn2+ because the d orbitals are completely filled. This means that the correct electronic configuration of the ion is [Ar] 3d10.
Answer:
Explanation:
<em>Endothermi</em>c processes absorb energy. The final state contains more energy than the initial state.
Since ice absorbs heat energy <em>in the process of completely melting</em> this is an <em>endothermic</em> process.
The process involves two stages: 1) heating the ice up to the melting point, which is 0ºC, and 2) melting the ice.
1. Heating the ice from -15ºC to 0ºC
a) Formula: Q = m×C×ΔT
- C = 2.108 kJ/kg.ºC (specific heat of ice)
b) Calculations:
- Q = m×C×ΔT = 1.6 kg × 2.108 kJ/kg.ºC × 15ºC = 50.592 kJ
2. Melting the ice at 0ºC
a) Formula: L = m × ΔHf
- ΔHf = 334 kJ/kg (latent heat of fussion)
b) Calculations
- L = m × ΔHf = 1.6 kg × 334 kJ/kg = 534.40 kJ
<u />
<u>2. Total heat</u>
<u />
- 50.592 kJ + 534.40 kJ = 584.992 kJ ≈ 590 kJ (rounded to 2 significant figures)
