2.083 Liters of 6.0 M solution sulfuric acid is required. This solved using molecular calculations and Titration.
Solution: 
Moles of hydrogen gas = 
Then 12.5 moles of hydrogen will be obtained from Moles of Sulfuric acid = 12.5 mol
Molarity of the sulfuric acid solution = 6.0 M = 6 mol/ l
6M = 
where V is the volume needed

V = 2.083 l
<h3>
What is Titration?</h3>
- Titration, commonly referred to as titrimetry, is a typical quantitative chemical analysis method used in laboratories to ascertain the unidentified quantity of an analyte .
- Titration is frequently referred to as volumetric analysis because it relies heavily on volume measurements. The titrant or titrator is a reagent that is prepared as a standard solution.
- To determine concentration, a solution of the analyte or titrand reacts with a known concentration and volume of the titrant. The titration volume is the amount of titrant that has responded.
- Titrations come in a variety of forms with various protocols and objectives. Redox and acid-base titrations are the two most typical types of qualitative titrations.
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Answer:
Uh first of all this is algebra but I'll answer this
First distribute the three and 5 (Multiply them by both terms inside parenthesis.
3x-6=5x+20
Then add like terms
8x=14
Divide 8 by 8 and 8 by 14
x = 14/8
Explanation:
Answer: It is a molecular compound. (2) It contains a metal. (3) It can conduct electricity as a solid.
When utilizing the gravimetric method, it is crucial to completely dissolve your sample in 10 mL of water. A quantitative technique called gravimetric analysis employs the selective precipitation of the component under study from an aqueous solution.
A group of techniques known as gravimetric analysis are employed in analytical chemistry to quantify an analyte based on its mass. Gravimetric analysis is a quantitative chemical analysis technique that transforms the desired ingredient into a substance (of known composition) that can be extracted from the sample and weighed. This is a crucial point to remember.
Gravimetric water content (g) is therefore defined as the mass of water per mass of dry soil. To calculate it, weigh a sample of wet soil, dry it to remove the water, and then weigh the dried soil (mdry). Dimensions of the sample Water is commonly forgotten despite having a density close to one.
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