For the answer to the question above, well presumably because the exact concentration of the composition KMnO4 solution doesn't matter. <span>If the concentration of the KMnO4 solution is important (usually in titrations etc.) then it is not allowed to use a wet bottle. The water in the bottle will dilute the KMnO4 solution and change the concentration of the said compound.</span>
Answer:
47.9 g of ethanol
Explanation:
Combustion is a chemical reaction in which a substance reacts with oxygen to produce heat and light. Combustion reactions have been very useful as a source of energy. Ethanol is now burnt for energy purposes as a fuel. Ethanol has even been proposed as a possible alternative to fossil fuels.
Since 1 mole of ethanol when combusted releases 1367 kJ/mol of energy
x moles of ethanol releases 1418 kJ/mol.
x= 1 × 1418 kJ/mol/ 1367 kJ/mol
x= 1.04 moles of ethanol.
Mass of ethanol = number of moles × molar mass
Molar mass of ethanol = 46.07 g/mol
Mass of ethanol = 1.04 moles × 46.07 g/mol
Mass of ethanol= 47.9 g of ethanol
The Patch's area of the space shuttle in km² is 2.07 × 10⁻⁹ km²
Given, that a space shuttle requires a 20.7 cm² patch
We have to convert the patch's area from cm² into km².
Unit conversion is a method in which we multiply or divide with a particular numerical factor and then finally round off to the nearest significant digits.
Patch area of the space shuttle is 20.7 cm²
1 cm = 0.00001 km
or, 1 cm² = (0.00001 km)²
or, 1 cm² = 10⁻¹⁰km²
20.7 cm² = 20.7 × 10⁻¹⁰km²
20.7 cm² = 2.07 × 10⁻⁹ km²
The patch area in square kilometers is 2.07 × 10⁻⁹ km²
To learn more about unit conversion, visit: brainly.com/question/11543684
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Answer:
50.8 g
Explanation:
Equation of reaction.
From the given information, the number of moles of methane = mass/ molar mass
= 15.4 g / 16.04 g/mol
= 0.960 mol
number of moles of oxygen gas = 90.3 g / 32 g/ mol
= 2.82 mol
Since 1 mol of methane requires 2 moles of oxygen
Then 0.960 mol of methane will require = 0.960 mol × 2 = 1.92 mol of oxygen gas
Thus, methane serves as a limiting reagent.
2.82 mol oxygen gas will result in 2.82 moles of water
So, the theoretical yield of water = moles × molar mass
= 2.82 mol × 18.01528 g/mol
= 50.8 g
<h3><u>Answer</u>;</h3>
≈ 4.95 g/L
<h3><u>Explanation;</u></h3>
The molar mass of KCl = 74.5 g/mole
Therefore; 0.140 moles will be equivalent to ;
= 0.140 moles × 74.5 g/mole
= 10.43 g
Concentration in g/L
= mass in g/volume in L
= 10.43/2.1
= 4.9667
<h3> <u> ≈ 4.95 g/L</u></h3>
