Answer:
Explanation:
Equation of the reaction:
NaOH + HCl --> NaCl + H2O
Volume of HCl = 5 ml
Molar concentration = 1 M
Number of moles = molar concentration * volume
= 1 * 0.005
= 0.005 mol of HCl
By stoichiometry, 1 mole of HCl completely neutralizes 1 mole of NaOH
Therefore, number of moles of NaOH = 0.005 mol
Molar mass of NaOH = 23 + 16 + 1
= 40 g/mol
NaOH --> Na+ + OH-
Mass = molar mass * number of moles
= 40 * 0.005
= 0.2 g of Na+
Answer:
Final concentration of C at the end of the interval of 3s if its initial concentration was 3.0 M, is 3.06 M and if the initial concentration was 3.960 M, the concentration at the end of the interval is 4.02 M
Explanation:
4A + 3B ------> C + 2D
In the 3s interval, the rate of change of the reactant A is given as -0.08 M/s
The amount of A that has reacted at the end of 3 seconds will be
0.08 × 3 = 0.24 M
Assuming the volume of reacting vessel is constant, we can use number of moles and concentration in mol/L interchangeably in the stoichiometric balance.
From the chemical reaction,
4 moles of A gives 1 mole of C
0.24 M of reacted A will form (0.24 × 1)/4 M of C
Amount of C formed at the end of the 3s interval = 0.06 M
If the initial concentration of C was 3 M, the new concentration of C would be (3 + 0.06) = 3.06 M.
If the initial concentration of C was 3.96 M, the new concentration of C would be (3.96 + 0.06) = 4.02 M
Answer:
Explanation:
Hello!
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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