<h2>Answer:</h2>
<h2>Explanations</h2>
The complete balanced equation for the given reaction is expressed as;
Given the following parameters
Mass of CH4 = 5.90×10^−3 g = 0.0059grams
Determine the moles of methane
According to stoichimetry, 1 mole of methane produces 2 moles of water, hence the moles of water required will be:
Determine the mass of water produced
Therefore the mass of water produced from the complete combustion of 5.90×10−3 g of methane is 1.33 * 10^-2grams
Answer:
Your answer is :- C [OH-] = 10 x 10-7 mol dm-3
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Answer:
This part require data such as Avogadro's number and the molar mass of water. But first, let's find the mass of water in the specified volume by making use of the density formula:
Density = mass/volume
1 g/mL = Mass/70 mL
Mass = 70 g
Each water contains 18 grams per mole, and each mole contains 6.022×10²³ molecules of water. Thus,
70 g * 1mole/18 g * 6.022×10²³ molecules/mole = 2.342×10²⁴ molecules of water
Explanation:
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