Yes u can but u have to divide mass by volum
<span>The Avogadro constant, named after the scientist Amedeo Avogadro, is the number of constituent particles, usually atoms or molecules, that are contained in the amount of substance given by one mole. The answer, in this case, would be 0.0004999...*10^23 = (10^19)*(4.999... [any finite number of 9s]).</span>
Answer:
The molar concentration of Cu²⁺ in the initial solution is 6.964x10⁻⁴ M.
Explanation:
The first step to solving this problem is calculating the number of moles of Cu(NO₃)₂ added to the solution:
n = 1.375x10⁻⁵ mol
The second step is relating the number of moles to the signal. We know the the n calculated before is equivalent to a signal increase of 19.9 units (45.1-25.2):
1.375x10⁻⁵ mol _________ 19.9 units
x _________ 25.2 units
x = 1.741x10⁻⁵mol
Finally, we can calculate the Cu²⁺ concentration :
C = 1.741x10⁻⁵mol / 0.025 L
C = 6.964x10⁻⁴ M
Answer:
Percent by mass of water is 56%
Explanation:
First of all calculate the mass of hydrated compound as,
Mass of Sodium = Na × 2 = 22.99 × 1 = 45.98 g
Mass of Sulfur = S × 1 = 32.06 × 1 = 32.06 g
Mass of Oxygen = O × 14 = 16 × 14 = 224 g
Mass of Hydrogen = H × 20 = 1.01 × 20 = 20.2 g
Mass of Na₂S0₄.10H₂O = 322.24 g
Secondly, calculate mass of water present in hydrated compound. For this one should look for the coefficient present before H₂O in molecular formula of hydrated compound. In this case the coefficient is 10, so the mass of water is...
Mass of water = 10 × 18.02
Mass of water = 180.2 g
Now, we will apply following formula to find percent of water in hydrated compound,
%H₂O = Mass of H₂O / Mass of Hydrated Compound × 100
Putting values,
%H₂O = 180.2 g / 322.24 g × 100
%H₂O = 55.92 % ≈ 56%
Answer:
The amount of HC₂H3₃2(aq) in the flask after the addition of 5.0mL of NaOH(aq) compared to the amount of HC₂H₃O₂(aq) in the flask after the addition of 1.0mL is much smaller because more HC₂H₃O₂(aq) is required to react with 5.0 mL NaOH than with 1.0 mL NaOH.
Explanation:
Equation of the reaction between acetic acid, HC₂H₃O₂(aq) and sodium hydroxide, NaOH(aq) is given below:
CH₃COOH (aq) + NaOH (aq) ----> CH₃COONa (aq) + H₂O
The equation of the reaction shows that acetic acid andsodium hydroxide will react in a 1:1 ratio
Since the concentration of NaOH was not given, we can assume that the concentration is 0.01 M
Moles of NaOH in 5.0 mL of 0.01 M NaOH = 0.01 × 5/1000 = 0.00005 moles
Moles of NaOH in 1.0 mL of 0.01 M NaOH = 0.01 ×1/1000 = 0.0001 moles
Ratio of moles of NaOH in 5.0 mL to 1.0 mL = 0.00005/0.00001 = 5
There are five times more moles of NaOH in 5.0 mL than in 1.0 mL and this means that 5 times more the quantity of HC₂H₃O2(aq) required to react with 1.0 mL NaoH is needed to react with 5.0 mL NaOH.
Therefore, the amount of HC₂H₃O2(aq) in the flask after the addition of 5.0mL of NaOH(aq) compared to the amount of HC₂H₃O₂(aq) in the flask after the addition of 1.0mL is much smaller because more HC₂H₃O₂(aq) is required to react with 5.0 mL NaOH than with 1.0 mL NaOH.
