Answer: The balanced reaction is:
<span>2 C6H14(g)+ 19 O2(g) → 12 <span>CO2</span>(g)+ 14 H2O(g)
Note: While balancing the chemical reaction, care must be taken that total number atoms (of each type) on both reactant and product side must be same. In present case, there are 12 'C' atoms, 28 'H' atoms and 38 'O' atoms on both reactant and product side. Hence, the reaction is balanced. </span>
Answer:
Explanation:
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In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:
Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:
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H^+(aq) + OH^-(aq) ---> H2O(l)
<span>Na^+ and ClO4^- are the spectator ions.</span>
It is more likely 9. pH 4 is acidic and pH 9 is basic, and as the pH of a substance gets closer to 0 or 14, the substance becomes more corrosive or reactive. As 4 is closer to 0 than 9 is to 14, there is a much higher chance the solution has a pH of 9, because pH 4 is less neutral and therefore more corrosive/reactive than pH 9.
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
