Number of moles of FeCl2 used = mass/ molar mass
Number of moles = 507/126.751 = 4.
If one mole of Fe reacts with two moles of sodium
Then 4 moles of Fe produces 8 moles of sodium.
Number of moles of sodium = mass/molar mass
Molar mass of sodium chloride = 23 +35.5 = 58.5 g/mol
Hence mass = 8 * 58.5 = 468 g. Hence Option A.
Answer:
Explanation:
Hello,
Based on the given undergoing chemical reaction is is rewritten below:
By stoichiometry we find the minimum mass of H2SO4 (in g) as shown below:
Moreover, mass of H2 gas (in g) would be produced by the complete reaction of the aluminum block turns out:
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Answer:
Polarizing power refers to an atoms ability to pull an electron toward it, polarizing the atom the electron comes from. Since cations are positive, they are able to attract electrons toward themselves. Anions are negative and so do not attract more electrons.
Therefore, Be2+ has a higher polarizing power because it has a higher quantitiy of protons, hence a higher polarizing power.
The number of mole of HCl needed for the solution is 1.035×10¯³ mole
<h3>How to determine the pKa</h3>
We'll begin by calculating the pKa of the solution. This can be obtained as follow:
- Equilibrium constant (Ka) = 2.3×10¯⁵
- pKa =?
pKa = –Log Ka
pKa = –Log 2.3×10¯⁵
pKa = 4.64
<h3>How to determine the molarity of HCl </h3>
- pKa = 4.64
- pH = 6.5
- Molarity of salt [NaZ] = 0.5 M
- Molarity of HCl [HCl] =?
pH = pKa + Log[salt]/[acid]
6.5 = 4.64 + Log[0.5]/[HCl]
Collect like terms
6.5 – 4.64 = Log[0.5]/[HCl]
1.86 = Log[0.5]/[HCl]
Take the anti-log
0.5 / [HCl] = anti-log 1.86
0.5 / [HCl] = 72.44
Cross multiply
0.5 = [HCl] × 72.44
Divide both side by 72.44
[HCl] = 0.5 / 72.4
[HCl] = 0.0069 M
<h3>How to determine the mole of HCl </h3>
- Molarity of HCl = 0.0069 M
- Volume = 150 mL = 150 / 1000 = 0.15 L
Mole = Molarity x Volume
Mole of HCl = 0.0069 × 0.15
Mole of HCl = 1.035×10¯³ mole
<h3>Complete question</h3>
How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (Ka of HZ is 2.3 x 10 -5 .) Assume negligible volume of the HCl
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