Answer:
0.05 mol
Explanation:
The balanced equation for the reaction that takes place is:
- 2C₂H₂ (g) + 5O₂ (g) → 4CO₂ (g) + 2H₂O (g)
Now we<u> convert 0.10 moles of carbon dioxide (CO₂) into moles of acetylene (C₂H₂)</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:
- 0.10 mol CO₂ *
= 0.05 mol C₂H₂
I believe it’s A, they form whole-number ratios in the compound
The density of the liquid at the given temperature is 1.622 g/mL
<h3>What is density? </h3>
The density of a substance is simply defined as the mass of the subtance per unit volume of the substance. Mathematically, it can be expressed as
Density = mass / volume
<h3>How to determine the density </h3>
The density of the liquid can be obtained as illustrated below:
- Mass of liquid = 40.55 g
- Volume of liquid = 25 mL
- Density of liquid =?
Density = mass / volume
Density = 40.55 / 25
Density of liquid = 1.622 g/mL
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Answer #1. A 2.5% (by mass) solution concentration signifies that there is 2.5 grams of solute in every 100 g of solution.
To calculate 2.5% by mass solution, we divide the mass of the solute by the mass of the solution and then multiply by 100.
Answer #2. therefore, when 2.5% is expressed as a ratio of solute mass over solution mass, that mass ratio would be 2.5/100 or 2.5 grams of solute/100 grams of solution.
This means that weighing out 2.5 grams of solute and then adding 97.5 grams of solvent would make a total of 100 gram solution:
mass of solute / mass of solution = 2.5g solute / (2.5g solute + 97.5g solvent)
= 2.5g solute / 100g solution
Answer#3. a solution mass of 1 kg is 10 times greater than 100 g, thus 1kg of a 2.5% ki solution would contain 25 grams of ki.
Since 1000 grams is 1 kg, we multiply 10 to each mass so that 100 grams becomes
1000grams:
mass of solute / mass of solution = 2.5g*10 / [(2.5g*10) + (97.5g*10)]
= 25g solute/(25g solute + 975g solvent)
= 25g solute/1000g solution
= 25g solute/1kg solution
The volume of 15.7 M H2SO4 is required to prepare 12.0 L of 0.156 M sulfuric acid is 0.12 L
<h3>Data obtained from the question</h3>
From the question given above, the following data were obtained:
- Molarity of stock solution (M₁) = 15.7 M
- Volume of diluted solution (V₂) = 12 L
- Molarity of diluted solution (M₂) = 0.156 M
- Volume of stock solution needed (V₁) = ?
<h3>How to determine the volume of the stock solution needed</h3>
The volume of the stock solution needed can be obtained by using the dilution formula as shown below:
M₁V₁ = M₂V₂
15.7 × V₁ = 0.156 × 12
15.7 × V₁ = 1.872
Divide both side by 15.7
V₁ = 1.872 / 15.7
V₁ = 0.12 L
Thus, the volume of the stock solution needed to prepare the solution is 0.12 L
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