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Lelu [443]
4 years ago
7

A chemist measures the amount of iodine solid produced during an experiment. He finds that of iodine solid is produced. Calculat

e the number of moles of iodine solid produced. Round your answer to significant digits.
Chemistry
1 answer:
Dimas [21]4 years ago
8 0

The question is incomplete, the complete question is:

A chemist measures the amount of iodine solid produced during an experiment. He finds that 8.31 g of iodine solid is produced. Calculate the number of moles of iodine solid produced. Round your answer to the correct number of significant digits.

<u>Answer: </u>The number of moles of solid iodine produced is 0.0327 moles

<u>Explanation:</u>

The number of moles is defined as the ratio of the mass of a substance to its molar mass.  The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

Given mass of solid iodine = 8.31 g

Molar mass of solid iodine = 253.8089 g/mol

Plugging values in equation 1:

\text{Moles of solid iodine}=\frac{8.31g}{253.8089g/mol}=0.0327mol

Hence, the number of moles of solid iodine produced is 0.0327 moles

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A phosphate buffer solution (25.00 mL sample) used for a growth medium was titrated with 0.1000 M hydrochloric acid. The compone
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Answer:

0,07448M of phosphate buffer

Explanation:

sodium monohydrogenphosphate (Na₂HP) and sodium dihydrogenphosphate (NaH₂P) react with HCl thus:

Na₂HP + HCl ⇄ NaH₂P + NaCl <em>(1)</em>

NaH₂P + HCl ⇄ H₃P + NaCl <em>(2)</em>

The first endpoint is due the reaction (1), When all phosphate buffer is as NaH₂P form, begins the second reaction. That means that the second endpoint is due the total concentration of phosphate that is obtained thus:

0,01862L of HCl×\frac{0,1000mol}{L}= 1,862x10⁻³moles of HCl ≡ moles of phosphate buffer.

The concentration is:

\frac{1,862x10^{-3}moles}{0,02500L} = <em>0,07448M of phosphate buffer</em>

<em></em>

I hope it helps!

7 0
4 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.

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