Answer:
Volume of acid, Va=250mL; Volume of quinine,Vb=20mL; Molarity of acid, Ma=0.05M.
Molar mass of acid= H2+S+O4= 2+32+(16X4)= 2+32+64=98g
Concentration of acid, Ca= Molar mass of acid/ Ma =98/0.05=1960g/mol
Explanation: To calculate concentration of quinine, Cb is as follow
Va*Ca=Vb*Cb
∴ Cb=Va*Ca/Vb =250*1960/20 =24500g/mol
Answer:
2C₃H₇BO₃ + 8O₂ → 6CO₂ + 7H₂O + B₂O₃.
Explanation:
- For balancing a chemical equation, we should apply the law of conversation of mass. It states that the no. of atoms in the reactants side is equal to that of the products side.
So, the balanced equation:
<em>2C₃H₇BO₃ + 8O₂ → 6CO₂ + 7H₂O + B₂O₃.</em>
It is clear that 2.0 moles of C₃H₇BO₃ is completely burned in 8 m oles of oxygen and produce 6 moles of CO₂, 7 moles of H₂O and 1 mole of B₂O₃.
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Answer : The radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.
Explanation :
As we are given that the Na⁺ radius is 56.4% of the Cl⁻ radius.
Let us assume that the radius of Cl⁻ be, (x) pm
So, the radius of Na⁺ = 
In the crystal structure of NaCl, 2 Cl⁻ ions present at the corner and 1 Na⁺ ion present at the edge of lattice.
Thus, the edge length is equal to the sum of 2 radius of Cl⁻ ion and 2 radius of Na⁺ ion.
Given:
Distance between Na⁺ nuclei = 566 pm
Thus, the relation will be:
The radius of Cl⁻ ion = (x) pm = 181 pm
The radius of Na⁺ ion = (0.564x) pm = (0.564 × 181) pm =102.084 pm ≈ 102 pm
Thus, the radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.
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