Answer:
Sodium Hydroxide
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Explanation:
Sodium hydroxide is a common proton acceptor also known as lye. It is the active ingredient in oven cleaners and drain cleaners because it can react with fats to produce compounds that dissolve in water. It is also used in many school laboratories to demonstrate acid-base reactions.
Answer:
The concentration the student should write down in her lab is 2.2 mol/L
Explanation:
Atomic mass of the elements are:
Na: 22.989 u
S: 32.065 u
O: 15.999 u
Molar mass of sodium thiosulfate, Na2S2O3 = (2*22.989 + 2*32.065 + 3*15.999) g/mol = 158.105 g/mol.
Mass of Na2S2O3 taken = (19.440 - 2.2) g = 17.240 g.
For mole(s) of Na2S2O3 = (mass taken)/(molar mass)
= (17.240 g)/(158.105 g/mol) = 0.1090 mole.
Volume of the solution = 50.29 mL = (50.29 mL)*(1 L)/(1000 mL)
= 0.05029 L.
To find the molar concentration of the sodium thiosulfate solution prepared we use the formula:
= (moles of sodium thiosulfate)/(volume of solution in L)
= (0.1090 mole)/(0.05029 L)
= 2.1674 mol/L
Explanation:
5g C7H8 / 92g/mole = 0.054moles
total moles = 0.232moles
mole fraction C6H12 = 0.178moles / 0.232moles = 0.767
mole fraction C7H8 = 0.233
Answer:
Explanation:
If an antacid has sodium hydrogen carbonate/Calcium carbonate, it reacts with HCl as shown
NaHCO₃+ HCl → NaCl + CO₂+ H₂O
Antacid acid salt gas water
CaCO₃+ 2HCl → CaCl₂+ CO₂+ H₂O
Antacid acid salt gas water
The formation of gas CO₂ is shown by brisk effervescence when the antacid (sodium hydrogen carbonate/calcium carbonate) reacts with HCl (acid). So CO₂ is the additional product formed and its formation is supported by observation of brisk effervescence as HCl is added to the antacid.
Answer:
83.64%.
Explanation:
∵ The percent yield = (actual yield/theoretical yield)*100.
actual yield of CO₂ = 2300 g.
- We need to find the theoretical yield of CO₂:
For the reaction:
<em>CH₄ + 2O₂ → 2H₂O + CO₂,</em>
1.0 mol of CH₄ react with 2 mol of O₂ to produce 2 mol of H₂O and 1.0 mol of CO₂.
- Firstly, we need to calculate the no. of moles of 1000 g of CH₄ using the relation:
<em>no. of moles of CH₄ = mass/molar mass</em> = (1000 g)/(16.0 g/mol) = <em>62.5 mol.</em>
<u><em>Using cross-multiplication:</em></u>
1.0 mol of CH₄ produces → 1.0 mol of CO₂, from stichiometry.
∴ 62.5 mol of CH₄ produces → 62.5 mol of CO₂.
- We can calculate the theoretical yield of carbon dioxide gas using the relation:
∴ The theoretical yield of CO₂ gas = n*molar mass = (62.5 mol)(44.0 g/mol) = 2750 g.
<em>∵ The percent yield = (actual yield/theoretical yield)*100.</em>
actual yield = 2300 g, theoretical yield = 2750 g.
<em>∴ the percent yield</em> = (2300 g/2750 g)*100 = <em>83.64%.</em>
