How many millimeters (mm) is the length of a standard table if it is

During a spectrophotometric analysis of an unknown, you measure your solution's absorbance, and the value is higher than the hig

Reil [10]

Answer:

A) Dilute the unknown so that it will have an absorbance within the standard curve. Once the diluted unknown concentration is determined, the full strength concentration can be calculated if the dilution process is recorded. Beer's law only applies to dilute solutions, so diluting the unknown is better than making new standards.

Explanation:

Beer's law states that <em>absorbance is proportional to the concentrations of the absorbing species</em>. This is verified in the case of diluted solutions (0≤0.01 M) of most substances. <u>As a solution gets more concentrated, solute molecules interact between themselves because of their proximity. </u>When a molecule interacts with another, the change in their electric properties (including absorbance) is probable. That's why <u>the plot of absorbance versus concentration stops being a straight line</u>, and <u>Beer's law is no longer valid.</u>

Therefore, if the absorbance value is higher than the highest standard, dilutions should be made. Once this concentration is determined, the full strength concentration can be calculated with the inverse of the dilution.

7 0

3 years ago

A compound is found to contain 50. 05% sulfur and 49. 95% oxygen by mass. What is the empirical formula for this compound? SO S2

jekas [21]

The empirical formula for the given compound has been $\rm SO_2$ . Thus, option C is correct.

The empirical formula has been the whole unit ratio of the elements in the formula unit.

<h3>Computation for the Empirical formula</h3>

The given mass of Sulfur has been, 50.05 g

The given mass of oxygen has been 49.95 g.

The moles of elements in the sample has been given by:

$\rm Moles=\dfrac{Mass}{Molar\;mass}$

Moles of Sulfur:

$\rm Moles\;S=\dfrac{50.05}{32}\\
 Moles\;S=1.56\;mol$

The moles of sulfur in the unit has been 1.56 mol.

Moles of Oxygen:

$\rm Moles\;O=\dfrac{49.95}{16} \\
Moles\;O=3.12\;mol$

The moles of oxygen in the unit has been 3.12 mol.

The empirical formula unit has been given as:

$\rm S_{1.56}O_{3.12}=SO_2$

Thus, the empirical formula for the given compound has been $\rm SO_2$ . Thus, option C is correct.

Learn more about empirical formula, here:

brainly.com/question/11588623

8 0

2 years ago

Consider the reaction between reactants s and o2: 2s(s)+3o2(g)→2so3(g)if a reaction vessel initially contains 7 mols and 9 mol o

nadya68 [22]

Answer: -

1 mol

Explanation: -

Number of moles of Sulphur S = 7

Number of moles of O2 = 9

The balanced chemical equation for the reaction is

2S (s)+3 O2 (g)→2SO3(g)

From the above reaction we can see that

3 mol of O2 react with 2 mol of S

9 mol of O2 will react with $\frac{2 mol S x 9 mol O2}{3 mol O2}$

= 6 mol of S

Unreacted S = 7 - = 1 mol.

If a reaction vessel initially contains 7 mol S and 9 mol O2

1 mole of s will be in the reaction vessel once the reactants have reacted as much as possible

6 0

4 years ago

C3H8(g) + 5O2(g) ⟶ 3CO2(g) + 4H2O(g) H = -2220 kJ If 865.9 g of H2O is produced during this combustion, how much heat is generat

dem82 [27]

Answer:

3 × 10⁴ kJ

Explanation:

Step 1: Write the balanced thermochemical equation

C₃H₈(g) + 5 O₂(g) ⟶ 3 CO₂(g) + 4 H₂O(g) ΔH = -2220 kJ

Step 2: Calculate the moles corresponding to 865.9 g of H₂O

The molar mass of H₂O is 18.02 g/mol.

865.9 g × 1 mol/18.02 g = 48.05 mol

Step 3: Calculate the heat produced when 48.05 moles of H₂O are produced

According to the thermochemical equation, 2220 kJ of heat are evolved when 4 moles of H₂O are produced.

48.05 mol × 2220 kJ/4 mol = 2.667 × 10⁴ kJ ≈ 3 × 10⁴ kJ

5 0

3 years ago

0.53g of acetanilide was subjected to kjeldahl determination and the ammonia produced was collected in 50cm3 of 0.50M of h2so4.o

Lady bird [3.3K]

Answer:

10.57% of N in acetanilide

Explanation:

All nitrogen in the sample is converted in NH₃ in the Kjeldahl determination. The NH₃ reacts with H₂SO₄ as follows:

2NH₃ + H₂SO₄ → 2NH₄⁺ + SO₄²⁻

The acid in excess in titrated with Na₂CO₃ as follows:

Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂

To solve this question we must find the moles of sodium carbonate = Moles of H₂SO₄ in excess. The added moles - Moles in excess = Moles of sulfuric acid that reacts:

<em>Moles Na₂CO₃ anf Moles H₂SO₄ in excess:</em>

0.025L * (0.05mol / L) = 1.25x10⁻³ moles Na₂CO₃ / 0.01360L =

0.09191M * 0.250L = 0.0230 moles H₂SO₄ in excess.

<em>Moles H₂SO₄ added:</em>

0.050L * (0.50mol / L) = 0.0250 moles H₂SO₄ added

<em>Moles that react:</em>

0.0250 moles - 0.0230 moles = 0.0020 moles H₂SO₄

<em>Moles of NH₃ = Moles N:</em>

0.0020 moles H₂SO₄ * (2mol NH₃ / 1mol H₂SO₄) = 0.0040 moles NH₃ = Moles N

<em>mass N and mass percent:</em>

0.0040 moles N * (14g / mol) = 0.056gN / 0.53g * 100 =

<h3>10.57% of N in acetanilide</h3>

7 0

3 years ago