SHOW WORK

When the concentration of hydronium ions equals the concentration of hydroxide ions, it is called the _____. balence?

GrogVix [38]

The answer is Equivalence point. When the condition of [H+] = [OH-] is reached, this state is called the equivalence point.

6 0

3 years ago

A solution is prepared by dissolving 33.0 milligrams of sodium chloride in 1000. L of water. Assume a final volume of 1000. lite

zheka24 [161]

Answer:

a. Molarity of NaCl solution = 5.64 * 10⁻⁷ mol/L

b. molarity of Na⁺ = 5.64 * 10⁻⁷ mol/L

c. molarity of Cl⁻ = 5.64 * 10⁻⁷ mol/L

d. Osmolarity = 1.128 osmol

e. mass percent of NaCl = 3.30 * 10⁻⁶ %

f. parts per million NaCl = 0.033 ppm NaCl

g. parts per billion of NaCl = 33 ppb of NaCl

h. From the values obtained from e, f and g, the most convenient to use and understand is parts per billion as it has less of a fractional part to deal with especially since the solute concentration is very small.

Explanation:

Molarity of a solution = number of moles of solute (moles)/volume of solution (L)

where number of moles of solute = mass of solute (g)/molar mass of solute (g/mol)

a. Molarity of NaCl:

molar mass of NaCl = 58.5 g/mol, mass of NaCl = 33.0/1000) g = 0.033g

number of moles of NaCl = 0.033/58.5 = 0.000564 moles

Molarity of NaCl solution = 0.000564/1000 = 5.64 * 10⁻⁷ mol/L

b. Equation for the dissociation of NaCl in solution: NaCl ----> Na⁺ + Cl⁻

From the above equation I mole of NaCl dissociates to give 1 mole of Na⁺ ions,

Therefore molarity of Na⁺ = 1 * 5.64 * 10⁻⁷ mol/L = 5.64 * 10⁻⁷ mol/L

c. From the above equation I mole of NaCl dissociates to give 1 mole of Cl⁻ ions,

therefore molarity of Cl⁻ = 1 * 5.64 * 10⁻⁷ mol/L = 5.64 * 10⁻⁷ mol/L

d. From the above equation, dissociation of NaCl in water produces 1 mol Na⁺ and 1 mole Cl⁻.

Total number of particles produced = 2

Osmolarity of solution = number of particles * molarity of siolution

Osmolarity = 2 * 5.64 * 10⁻⁷ mol/L = 1.128 osmol

e. mass of percent of NaCl = {mass of NaCl (g)/ mass of solution (g)} * 100

density of water = 1 Kg/L

mass of water = 1 Kg/L * 1000 L = 1000 kg

1Kg = 1000 g

Therefore mass of solution in g = 1000 * 1000 = 1 * 10⁶ g

mass percent of NaCl = (0.033/1 * 10⁶) * 100 = 3.30 * 10⁻⁶ %

f. Parts per million of NaCl:

parts per million = 1 mg of solute/L of solution

One thousandth of a gram is one milligram and 1000 ml is one liter, so that 1 ppm = 1 mg per liter = mg/Liter.

Since the density of water is 1kg/L = 1,000,000 mg/L

1mg/L = 1mg/1,000,000mg or one part in one million.

parts per million NaCl = 33.0/1000 L = 0.033 ppm NaCl

g. Parts per billion = 1 µg/L of solution

1 g = 1000 µg

therefore, 33.0 mg = 33.0 * 1000 µg = 3.30 * 10⁴ µg

parts per billion of NaCl = 3.30 * 10⁴ µg/1000 L = 33 ppb of NaCl

h. From the values obtained from e, f and g, the most convenient to use and understand is parts per billion as it has less of a fractional part to deal with especially since the solute concentration is very small.

5 0

3 years ago

NaClO3 > NaCl + O2 Balance

tiny-mole [99]

Answer:

Balancing Strategies: To balance this reaction it is best to get the Oxygen atoms on the reactant side of the equation to an even number. Once this is done everything else falls into place. Put a "2" in front of the NaClO3. Change the coefficient in front of the O2.

6 0

3 years ago

One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of silv

babunello [35]

Answer:

5.0x10⁻⁵ M

Explanation:

It seems the question is incomplete, however this is the data that has been found in a web search:

" One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of silver nitrate solution. Any chloride anions in solution will combine with the silver cations to produce bright white silver chloride precipitate. Suppose a EPA chemist tests a 250 mL sample of groundwater known to be contaminated with nickel(II) chloride, which would react with silver nitrate solution like this:

NiCl₂ + 2AgNO₃ → 2AgCl + Ni(NO₃)₂

The chemist adds 50 mM silver nitrate solution to the sample until silver chloride stops forming. She then washes, dries, and weighs the precipitate. She finds she has collected 3.6 mg of silver chloride. Calculate the concentration of nickel(II) chloride contaminant in the original groundwater sample. Round your answer to 2 significant digits. "

Keep in mind that while the process is the same, if the values in your question are different, then your answer will be different as well.

First we calculate the moles of nickel chloride found in the 250 mL sample:

3.6 mg AgCl ÷ 143.32 mg/mmol * $\frac{1mmolNiCl_{2}}{2mmolAgCl}$ = 0.0126 mmol NiCl₂

Now we divide the moles by the volume to calculate the molarity:

0.0126 mmol / 250 mL = 5.0x10⁻⁵M

4 0

3 years ago

Describe three devices used in radiation detection

xeze [42]

MicroR Meter, with Sodium Iodide Detector
Geiger Counter, with Geiger-Mueller (GM) Tube or Probe
Portable Multichannel Analyzer

6 0

3 years ago