A mixture has lots of different elements that are not necessarily bonded to each other, like sea water has lots of dirt, animals, and plant parts in it. Compared to a solution (strictly salt and water, which bond and ionize with each other).
The correct answer is 0.15.
We are aware that there is 0.05 mol of an unidentified hydrocarbon we will refer to as "X" and that its burning produces 6.6 g of carbon dioxide and 3.6 g of water.
These quantities might be converted to moles by applying the following formula:
amount= mass/ relative atomic mass
Thus, the following equation may be written for H2O: moles = 3.6 / 18 = 0.2 and for CO2: moles = 6.6 / 44 = 0.15.
0.05X + x'O2 = 0.15CO2 + 0.2H2O
This may be made simpler by dividing through by 0.05 (this step is likely to be the most helpful to you), resulting in:
1 x + x O2 = 3 co2 + 4 H2O
The hydrocarbon must have been the source of all the carbon in the carbon dioxide and all the hydrogen in the water.
Accordingly, 4 x 2 = 8 moles of H and 3 x 1 = 3 moles of C.
There are 3/1 = 3 Cs and 8/1 = 8 Hs in one X molecule.
This clearly identifies C3H8 or propane as the hydrocarbon X (dividing by 1 seems unnecessary, but it illustrates the process to use if there were more than one mol of X in the first equation).
To learn more about number of moles of carbon dioxide refer the link:
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Answer:
Precipitation reaction
Explanation:
Given that solution A was mixed with solution B, the solution turned cloudy. The test is not warm and no bubbles visible. This means that the precipiate is formed.
The concept is when two colourless solutions react to form a cloudy precipitate that settles at bottom of a solution then the reaction is said to be a precipitation reaction.
An example can be the Reaction of Silver nitrate with common salt.
Answer:
- colorless
- noble gases are arranged according to their boiling point in the periodic table
- tasteless
- non flammable in stander conditions
Answer:
364 K or 91°C
Explanation:
Applying,
V₁/T₁ = V₂/T₂................ Equation 1
Where V₁ = Initial Volume, V₂ = Final volume, T₁ = initial Temperature, T₂ = final Temperature.
make T₂ the subject of the equation,
T₂ = V₂T₁/V₁................. Equation 2
From the question,
Given: V₁ = 375 mL, V₂ = 500 mL, T₁ = 0.0°C = (273+0) K = 273 K
Substitute these values into equation 2
T₂ = (500×273)/375
T₂ = 364 K
T₂ = (364-273) °C = 91 °C
