The mass of NaCl needed for the reaction is 91.61 g
We'll begin by calculating the number of mole of F₂ that reacted.
- Gas constant (R) = 0.0821 atm.L/Kmol
PV = nRT
1.5 × 12 = n × 0.0821 × 280
18 = n × 22.988
Divide both side by 22.988
n = 18 / 22.988
n = 0.783 mole
Next, we shall determine the mole of NaCl needed for the reaction.
F₂ + 2NaCl —> Cl₂ + 2NaF
From the balanced equation above,
1 mole of F₂ reacted with 2 moles of NaCl.
Therefore,
0.783 mole F₂ will react with = 0.783 × 2 = 1.566 moles of NaCl.
Finally, we shall determine the mass of 1.566 moles of NaCl.
- Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass = mole × molar mass
Mass of NaCl = 1.566 × 58.5
Mass of NaCl = 91.61 g
Therefore, the mass of NaCl needed for the reaction is 91.61 g
Learn more about stiochoimetry: brainly.com/question/25830314
The initial mass of sodium hydroxide is 3.3 g (answer C)
<u><em>calculation</em></u>
Step 1 : find the moles of iron (ii) hydroxide ( Fe(OH)₂
moles = mass÷ molar mass
from periodic table the molar mass of Fe(OH)₂ = 56 + [16 +1]2 = 90 g/mol
moles is therefore = 3.70 g÷ 90 g/mol = 0.041 moles
Step 2: use the mole ratio to calculate the moles of sodium hydroxide (NaOH)
from given equation NaOH : Fe(OH)₂ is 2 :1
therefore the moles of NaOH = 0.041 x 2 = 0.082 moles
Step 3: find mass of NaOH
mass = moles x molar mass
from the periodic table the molar mass of NaOH = 23 +16 +1 = 40 g/mol
mass = 0.082 moles x 40 g/mol = 3.3 g ( answer C)
Answer:
C15 H31 O4 S
Explanation:
molecular formula is also the same because the value of "n" is 1
Answer: Option (A) is the correct answer.
Explanation:
When energy is transferred from the air to the water then energy is absorbed by the water molecules.
This energy travels through one molecule of water to another molecule of water by the process of convection.
Thus, we can conclude that when energy is transferred from the air to the water, then it travels through the water.
Answer:
a. 3; b. 5; c. 10; d. 12
Explanation:
pH is defined as the negative log of the hydronium concentration:
pH = -log[H₃O⁺] (hydronium concentration)
For problems a. and b., HCl and HNO₃ are strong acids. This means that all of the HCl and HNO₃ would ionize, producing hydronium (H₃O⁺) and the conjugate bases Cl⁻ and NO₃⁻ respectively. Further, since all of the strong acid ionizes, 1 x 10⁻³ M H₃O⁺ would be produced for a., and 1.0 x 10⁻⁵ M H₃O⁺ for b. Plugging in your calculator -log[1 x 10⁻³] and -log[1.0 x 10⁻⁵] would equal 3 and 5, respectively.
For problems c. and d. we are given a strong base rather than acid. In this case, we can calculate the pOH:
pOH = -log[OH⁻] (hydroxide concentration)
Strong bases similarly ionize to completion, producing [OH⁻] in the process; 1 x 10⁻⁴ M OH⁻ will be produced for c., and 1.0 x 10⁻² M OH⁻ produced for d. Taking the negative log of the hydroxide concentrations would yield a pOH of 4 for c. and a pOH of 2 for d.
Finally, to find the pH of c. and d., we can take the pOH and subtract it from 14, giving us 10 for c. and 12 for d.
(Subtracting from 14 is assuming we are at 25°C; 14, the sum of pH and pOH, changes at different temperatures.)
