Yes... that is correct.
CH4 is methane so the coefficent in front of it would double the number of atoms of each element
Given:
Concentration of Fluoride ions = 0.100 M
Concentration of Hydrogen Fluoride = 0.126 M
Asked: Concentration of fluoride ions after the addition of 5ml of 0.0100 M HCl to 25 mL of the solution
Assume: 50:50 ratio of fluoride ions and HF
12.5ml*0.1mol/L *1L/1000mL + 12.5*0.126mol/L * 1L/1000mL = 2.825x10^-3 moles F-
5ml * 0.01 mol/L *1L/1000mL = 5x10^-5 moles
Assume: Volume additive
Final concentration = 2.825x10^-3 + 5x10^-5 moles/ 30 ml * 1000ml/L =0.0958 M
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Answer: hello your question is incomplete below is the complete question
Salt water contains n sodium ions (Na+) per cubic meter and n chloride ions (Cl−) per cubic meter. A battery is connected to metal rods that dip into a narrow pipe full of salt water. The cross sectional area of the pipe is A. The magnitude of the drift velocity of the sodium ions is VNa and the magnitude of the drift velocity of the chloride ions is VCl.
What is the magnitude of the ammeter reading ?
answer :
| I | = neAVₙₐ + neAV (Cl-)
Explanation:
Given that there are N sodium ions
<u>Determine the Magnitude of the ammeter reading </u>
| I | = current due to sodium ions + current due to (Cl-) ions
= neAVₙₐ + neAV (Cl-)
I believe so..... hmm idk
