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Dennis_Churaev [7]
3 years ago
5

A forensic chemist is given a white powder for analysis. She dissolves 0.50 g of the substance in 8.0 g of benzene. The solution

freezes at 3.9°C. Can the chemist conclude that the compound is cocaine (C17H21N04)? What assumptions are made in the analysis? The freezing point of benzene is 5.5°C.
Chemistry
1 answer:
Sholpan [36]3 years ago
3 0

Answer:

The given compound cannot be cocaine.

Explanation:

The chemist can comment on the nature of compound being cocaine or not from the depression in freezing point.

Depression in freezing point of is related to molality as:

Depression in freezing point = Kf X molality

Where

Kf = cryoscopic constant = 4.90°C/m

depression in freezing point = normal freezing point - freezing point of solution

depression in freezing point = 5.5-3.9 = 1.6°C

1.6°C = 4.90 X molality

molality=\frac{1.6}{4.90} = 0.327 m

we know that:

molality=\frac{moles}{mass of solvent(kg)}=\frac{moles}{0.008kg}

therefore

moles = 0.327X0.008 = 0.00261 mol

moles=\frac{mass}{molarmass}

molarmass=\frac{mass}{moles}=\frac{0.50}{0.00261}= 191.57g/mol

The molar mass of cocaine is 303.353

So the given compound cannot be cocaine.

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NARA [144]

Answer:

5.8μg

Explanation:

According to the rate or decay law:

N/N₀ = exp(-λt)------------------------------- (1)

Where N = Current quantity,  μg

            N₀ = Original quantity, μg

             λ= Decay constant day⁻¹

              t =  time in days

Since the half life is 4.5 days, we can calculate the  λ from (1) by  substituting N/N₀ = 0.5

0.5 = exp (-4.5λ)

ln 0.5  = -4.5λ

-0.6931 = -4.5λ

λ =   -0.6931 /-4.5

  =0.1540 day⁻¹

Substituting into (1)  we have :

N/N₀ = exp(-0.154t)----------------------------- (2)

To receive 5.0 μg of the nuclide with a delivery time of 24 hours or 1 day:

N = 5.0 μg

N₀ = Unknown

t = 1 day

Substituting into (2) we have

[5/N₀]   = exp (-0.154 x 1)

    5/N₀        = 0.8572

N₀  =  5/0.8572

     =    5.8329μg

    ≈     5.8μg

The Chemist must order 5.8μg  of 47-CaCO3

6 0
3 years ago
What mass of HgO is required to produce 0.692 mol of O2?<br><br>2HgO(s) -&gt; 2Hg(l) + O2(g)
Vika [28.1K]

The answer for the following problem is mentioned below.

  • <u><em>Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule </em></u>

Explanation:

Given:

no of moles of the oxygen gas = 0.692

Also given:

2 HgO  → 2 Hg + O_{2}

where,

HgO represents mercuric oxide

Hg represents mercury

O_{2} represents oxygen

To calculate:

Molar mass of HgO:

Molar mass of HgO = 216 grams

molar mass of mercury (Hg) = 200 grams

molar mass of oxygen (O) =16 grams

HgO = 200 +16 = 216 grams

We know;

       2×216 grams of HgO   →  1 mole of oxygen molecule

             ?                              →  0.692 moles of oxygen molecule

       

          = \frac{2*216*0.692}{1}

      = 298.944 grams of HgO

<u><em>Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule </em></u>

<u />

7 0
3 years ago
Read 2 more answers
4. A student purified a 500-mg sample of phthalic acid by recrystallization from water. The published solubility of phthalic aci
kupik [55]

Answer:

2.77 mL of boiling water is the minimum amount which will dissolve 500 mg of phthalic acid.

Explanation:

We know from the problem that 18 g of phthalic acid are dissolved in 100 mL of water at 99 °C.

Now we devise the following reasoning:

If          18 g of phthalic acid are dissolved in 100 mL of water at 99 °C

Then   0.5 g of phthalic acid are dissolved in X mL of water at 99 °C

X = (0.5 × 100) / 18 = 2.77 mL of water

7 0
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Explanation:

Scientist use trees a whole lot to look at climate of the past by examining tree rings.

These are layers of cambium in each successive years formed. They have an annual growth pattern and are known as tree rings.

Tree rings can be used to decipher the age of a tree.

  • These three rings can be used to interpret climatic patterns.
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  • In a dry climate, the rings are thinner.
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6 0
3 years ago
Write a balanced equation for its combustion reaction.
laila [671]
<h3>Answer:</h3>

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<h3>Explanation:</h3>

The balanced chemical equation for the combustion of the hydrocarbon in question is;

C₅H₁₂O(l)+15/2O₂(g)→5CO₂(g)+6H₂O(l)

  • A balanced chemical equation is one in which the number of atoms of each element is equal on both sides of the equation.
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  • Product side has; 5 carbon atoms, 12 hydrogen atoms and 16 Oxygen atoms
  • An equation is balanced by putting appropriate coefficients on reactants and products involved in the reaction.
  • An equation is balanced so as to obey the law of conservation of mass.
5 0
3 years ago
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