Water-rich solutions carrying metallic salts fill cracks and form
1 answer:
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Answer:
Genotype of parents is CC and Cc
Explanation:
Please see the attachment
Answer:
0.005 M
Explanation:
Given data:
volume of sample solution ( volume of D ) = 5.0 mL
volume of added stock solution ( V1 ) = 5.0 mL
concentration of added stock solution ( N1 ) = 0.02 M
Total volume of concentration ( V2 )= 10 mL = ( 5.0 mL + 5.0mL)
concentration of Total volume of sample ( C2 ) = 0.01
N2 = ( N1V1 ) / V2
= ( 0.02 * 5 ) / 10 = 0.01 m
absorbance of sample solution ( A1 ) = 0.10
absorbance of additional sample solution ( A2 ) = 0.20
attached below is the remaining part of the detailed solution
Answer:
Here's what I get
Explanation:
3. Molar concentration by formula.
(i) Comparison of molar concentrations
The formula gives a calculated value of 0.5302 mol·L⁻¹.
Dimensional analysis gives a calculated value of 0.1767 mol·L⁻¹.
The first value is three times the second.
It is wrong because the formula assumes that the acid supplies just enough moles of H⁺ to neutralize the OH⁻ from the NaOH.
Instead, I mol of H₃PO₄ provides 3 mol of H⁺, so your calculated concentration is three times the true value.
(ii) When is the formula acceptable?
The formula is acceptable only when the molar ratio of acid to base is 1:1.
Examples are
HCl + NaOH ⟶ NaCl + H₂O
H₂SO₄ + Ca(OH)₂ ⟶ CaSO₄ + 2H₂O
H₃PO₄ + Al(OH)₃ ⟶ AlPO₄ + 3H₂O