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B) The elements become less reactive.
Answer:
41 mL
Explanation:
Given data:
Milliliter of HCl required = ?
Molarity of HCl solution = 4.25 M
Mass of CaCO₃ = 8.75 g
Solution:
Chemical equation:
2HCl + CaCO₃ → CaCl₂ + CO₂ + H₂O
Number of moles of CaCO₃:
Number of moles = mass/molar mass
Number of moles = 8.75 g / 100.1 g/mol
Number of moles = 0.087 g /mol
Now we will compare the moles of CaCO₃ with HCl.
CaCO₃ : HCl
1 : 2
0.087 : 2/1×0.087 = 0.174 mol
Volume of HCl:
Molarity = number of moles / volume in L
4.25 M = 0.174 mol / volume in L
Volume in L = 0.174 mol /4.25 M
Volume in L = 0.041 L
Volume in mL:
0.041 L×1000 mL/ 1L
41 mL
Answer:
c. chloroacetate ion
Explanation:
The chloroacetic acid, ClCH₂CO₂H, is a weak acid with Ka = 1.36x10⁻³. When this weak acid is in solution with its conjugate base, ClCH₂CO₂⁻ (From sodium chloroacetate) a buffer is produced. The addition of a strong acid as the HCl produce the following reaction
HCl + ClCH₂CO₂⁻ → ClCH₂CO₂H + Cl⁻.
Where the acid reacts with the chloroacetate ion to produce more chloroacetic acid
That means, the HCl reacts with the chloroacetate ion present in the buffer solution
Right answer is:
<h3>c. chloroacetate ion</h3>
